Unit-ll

Chapter 4

Principles of Mathematical induction

The Principle of Mathematical Induction

What is principle of mathematical induction ?

Mathematical Induction is a technique of proving a statement, theorem or formula which is thought to be true, for each and every natural number n. By generalizing this in form of a principle which we would use to prove any mathematical statement is 'Principle of Mathematical Induction'.

 

Steps:

1. Let the given statement be P(n)
2. The statement is true for n = 1 i.e., P(1) is true, and.
3. If the statement is true for n=k, where k is a positive integer, then the statement is also true for all cases of. n= k+1 i.e., P(k) leads to the truth of P(k+1).
4. Hence P(n) is true for all n Î N

Motivation and Process of proof by induction:

Mathematical Induction gets motivated by the real life examples like if we have a ladder to reach to the roof then how will you prove that you will reach at the top using these ladders.

• First thing we need to assure that can we reach to the first step of that ladder, as if we reach to the first step then only you can go to the second, third and so on. And if you will not reach to the first one then you cannot reach to the top.
• Second thing is if we count the steps of the ladder then if we assume that we can reach to the 4^th step then we can reach to the 5^th one also.

Let’s understand it in terms of numbers:

Let’s give the numbers to the ladders as number 1, 2, and so on. Let p (n) is the proposition to reach to the nth number of step. Now we have to prove that if we move to the any step then we can reach to the nth step.

This statement in mathematical term is the same as “for any k ≥ 0, if P (k) is true then P (k + 1) is also true.”

For this first we have to reach to the first step so in mathematical term, If P (1) is true, then only we can say that “if P (k) is true then P (k + 1) is also true” for k = 1 and then we get that P(2) is also true, which then indicates that P(3) is true, and so on, which shows that P(n) is true for all n ≥ 1. But if P (1) is not true, then P (n) also may not be true for any n ≥ 2.

Principle of Mathematical Induction

The above concept derives the principle of mathematical induction which is used to directly prove the statements given in terms of n natural numbers.

• Base Case: The given statement is correct for first n natural number that is, for n = 1, p (1) is true.
• Inductive Step: If the given statement is correct for any natural number like n = k then it will be correct for n = k + 1 also that is, if p (k) is true then p (k+1) is also true.

This principle says that if both the above steps are proven then p (n) is true for all natural numbers.

Example

Let’s take the example of the formula for the sum of the first n squares, for n ≥ 0. We will find a pattern in the sum of squares of n natural numbers:

0² = 0

0² + 1²  = 1

0² + 1² + 2² = 5

0² + 1² + 2² + 3² = 14

If we will follow this pattern we can see that the sum of the first n squares could be n (n + 1) (2n + 1)/6. Now we have to prove that this formula works for every number n ≥ 0, that is, we have to prove an infinite number of equalities. 

When can we use Mathematical Induction?

Mathematical induction can be used for so many statements where we want to show, that statement is true. Every statement cannot be proved by this new technique.

• First, in induction we need to find our statement P (n) for all integers n ≥ 0, so induction works only for the statements with the whole numbers. So, the statements like “For all x ∈ R, x² ≥ 0” will not be able to prove with  induction as the set of real numbers is not easy to calculate.
• Secondly, for our induction proof we have to follow the inductive step  that is, statements where the P (k+ 1) case is true assuming that the P (k) case is true are particularly well-suited for induction. For Example, statements involving sums  are suitable as it is easy to write the left-hand side of our statement P (k + 1) in terms of the left-hand side of the statement P (k).

Chapter 5

Complex Numbers and Quadratic Equations

Imaginary Numbers and Powers

It is a solution to the quadratic equation or expression, x²+1 = 0, such as;

x² = 0 – 1

x² = -1

x = √-1

x = i

Therefore, an imaginary number is the part of complex number which we can write like a real number multiplied by the imaginary unit i, where i² = -1. The imaginary number, when multiplied by itself, gives a negative value.

 

Value of Powers of i

We know, i² = -1, let us calculate the value of ‘i’ raised to the power other imaginary numbers.

i⁴ⁿ = 1
i⁴ⁿ⁺¹ = i
i⁴ⁿ⁺²= -1
i⁴ⁿ⁺³=-i

Chapter-6

Linear Inequalities

Inequalities and Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation:

What is inequalities

• In mathematics, an inequality is a relation that holds between two values when they are different
• Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative 
• Two real numbers or two algebraic expressions related by the symbol ‘<’, ‘>’, ‘£’ or ‘³’  form an inequality.

Symbols used in inequalities

• The symbol < means less than. The symbol > means greater than.
• The symbol < with a bar underneath means less than or equal to. Usually this is written as ≤
• The symbol > with a bar underneath means greater than or equal to. Usually this is written as ≥
• The symbol ≠ means the quantities on left and right side are not equal

Examples

• a < b means a is less then b or b is greater a
• a≤b means a is less then or equal to b
• a > b means a is greater than b
• a≥b means a is greater or equal to b

Types of inequalities:

1. Numerical inequalities :

Example:

3 < 5(read as 3 is less than 5)

7 > 5 (read as 7 is greater than 5)

2. Literal inequalities:

Inequalities which involve variables are called literal inequalities

Example: x < 5; y > 2

    x ³5; y £ 2;

3. Double inequalities:

The inequality is said to be a double inequality if the statement shows the double relation of the expressions or the numbers..

Example: 3 < 5 < 7 (read as 5 is greater than 3 and less than 7),

3 £ x < 5 (read as x is greater than or equal to 3 and less than 5)

And  2 < y £ 4(read as y is less than or equal to 4 and greater than 2)

4. Open Sentence :

The inequality is said to be an open sentence if it has only one variable.

Example: x < 6 (x is less than 6),

 x > 8

5. Strict inequalities:

A relation that expresses the comparison between the unequal quantities is called strict inequality.

Example: ax + b < 0,

ax + b > 0,

ax + by < c,

ax + by > c,

ax² + bx + c > 0

6. Slack inequalities:

 Inequalities involving the symbol '≥' or '≤' are called slack inequalities.

Example: ax + by +c,

ax + by +c,

3x – y ≥ 5

7. Linear inequalities:

When two expressions are connected by “greater than” or “less than” sign, we get an inequality. Also, the linear inequation is similar to a linear equation, where the equal to sign is replaced by the inequality sign.

Example: ax + b < 0,

ax + b > 0,

 x - 5 > 3x – 10,

8. Quadratic inequalities:

A quadratic inequality is an inequality that contains a quadratic expression.

Example:  ax²+bx+c<0,

ax² + bx + c > 0

ax² + bx + c £ 0

 

Things are which are safe to do in inequality which does not change in direction:

• Addition of same number on both sides
          a>b
  =>    a+c > b+c
• Subtraction of same number on both sides
        a>b
  =>  a−c > b−c
• Multiplication/Division by same positive number on both sides
      a>b
  if c is positive number then
    ac >  bc
  if c is positive number (non zero) then
    a/c >  b/c

Things which changes the direction of the inequality:

• Swapping the left and right sides
• Multiplication/Division by negative number on both sides

If a > b and  c is negative number then  ac <  bc

If a < b and  c is negative number then  ac > bc

• Don’t multiple by variable whose values you dont know as you don’t know the nature of the variable

Concept Of Number line:

• A number line is a horizontal line that has points which correspond to numbers. The points are spaced according to the value of the number they correspond to; in a number line containing only whole numbers or integers, the points are equally spaced.

  

• It is very useful in solving problem related to inequalities and also representing it
  Suppose x >2(1/ 3), this can represent this on number line like that

  

Linear Inequation in One Variable:

A inequation of the form
ax+b>0
or
ax+b≥0

or
ax+b<0
or
ax+b≤0
are called the linear equation in One Variable

      Example:

      x−2<0,

          3x+10>0,

10x−17≥0

Linear Inequations in Two Variable:

A inequation of the form
ax+by>c
or
ax+by≥c
or
ax+by<c
or
ax+by≤c
are called the linear equation in two Variable

Example:
 

• x−2y<0
• 3x+10y>0
• 10x−17y≥0

How to solve the linear inequalities in one variable?:

Steps to solve the Linear inequalities in one variable:

• Obtain the linear inequation
• Pull all the terms having variable on one side and all the constant term on another side of the inequation
• Simplify the equation in the form given above
  ax>b
  or
  ax≥b
  or
  ax<b
  or ax≤b
• Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed
• Put the result of this equation on number line and get the solution set in interval form

Example:
x−2 > 2x+15
Solution
x−2 > 2x+15
Subtract x from both the side x−2−x > 2x+15−x
−2>x+15

Subtract 15 from both the sides −17 > x
Solution set (−∞,−17)

Example: Find all pairs of consecutive odd natural numbers, both of which are larger

than 10, such that their sum is less than 40.

Solution: Let x be the smaller of the two consecutive odd natural number, so that the

other one is x +2. Then, we should have

x > 10 ... (1)

and x + ( x + 2) < 40 ... (2)

Solving (2), we get

2x + 2 < 40

i.e., x < 19 ... (3)

From (1) and (3), we get

10 < x < 19

Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required

possible pairs will be

(11, 13), (13, 15), (15, 17), (17, 19)

Steps to solve the linear inequality of the fraction form

[(ax+b)/(cx+d)]>k or similar type
 

• Take k on the LHS
• Simplify LHS to obtain the inequation in the form
  [(px+q)/(ex+f)]>0
  Make the coefficent positive if not
• Find out the end  points solving the equatoon px+q=0 and ex+f=0
• Plot these numbers on the Number line. This divide the number into three segment
• Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
• Write down the solution set in interval form

Or there is more method to solves these
 

• Take k on the LHS
• Simplify LHS to obtain the inequation in the form

[(px+q)/(ex+f)]>0
Make the coefficent  positive if not

•   For the equation to satisfy both the numerator and denominator must have the same sign
• So taking both the part +, find out the variable x interval
• So taking both the part -, find out the variable x interval
• Write down the solution set in interval form

Lets take one example to clarify the points

Lets take one example to clarify the points

Question

1. Solve     X−3x+5>0
Solution:
Method A
 

1. Lets find the end points of the equation
   Here it is clearly
   x=3 and x=-5
2. Now plots them on the Number line
3. Now lets start from left part of the most left number
   i.e
   Case 1
   x<−5 ,Let takes x=-6 then −6−3−6+5>0
   3>0
   So it is good
   Case 2
   Now take x=-5
   as x+5 becomes zero and we cannot have zero in denominator,it is not the solution
   Case 3
   Now x > -5 and x < 3, lets take x=1 then 1−31+5>0
   −16>0
   Which is not true
   Case 4
   Now take x =3,then
   0> 0 ,So this is also not true
   case 5
   x> 3 ,Lets x=4
   4−34+5>0
   19>0
   So this is good
4. So the solution is
   x < -5 or x > 3
   or
   (−∞,−5)∪(3,∞)

Method B
 

1. the numerator and denominator must have the same sign. Therefore, either
   1) x−3>0 and x+5>0,or 2) x - 3 < 0 and x + 5 < 0$
2. Now, 1) implies x > 3 and x > -5.
   Which numbers are these that are both greater than 3 and greater than -5?
   Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution
   x > 3.
3. Next, 2) implies
   x < 3 and x < -5.
   Which numbers are these that are both less than 3 and less than -5?
   Clearly, any number less than -5 will also be less than 3. Therefore, 2) has the solution
   x < -5.
4. The solution, therefore, is
   x < -5 or x > 3

 

Quadratic Inequality:

A inequation of the form
 so for a ≠ 0
ax²+bx+c>0
or
ax²+bx+c<0
or
ax²+bx+c≥0
or
ax²+bx+c≤0
are called Quadratic Inequation in one Variable

Steps to solve Quadratic or polynomial inequalities:

ax²+bx+c>0,
or
ax²+bx+c<0,
or
ax²+bx+c≥0,
ax²+bx+c≤0

1. Obtain the Quadratic inequation
2. Pull all the terms having on one side and Simplify the equation in the form given above
3. Find the roots(0 points) of the Quadratic equation using any of the method and write in this form
   (x-a)(x-b)
4. Plot these roots on the number line .This divide the number into three segment
5. Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
6. Write down the solution set in interval form

Solved Example
Question 1
x²-5x+6>0
Solution
1) Simplify or factorize the inequality which means factorizing the equation in case of quadratic equalities
Which can be simplified as
x²−5x+6>0
=> x²−2x−3x+6>0
=> (x−2)(x−3)>0
2) Now plot those points on Number line clearly
3) Now start from left of most left point on the Number line and look out the if inequalities looks good or not. Check for greater ,less than and equalities at all the end points
So in above case of  x²−5x+6>0
We have two ends points 2 ,3
Case 1
So for x<2 ,Let take x=1,then (1−2)(1−3)>0
=> 2>0
So it is good
So This inequalities is good for x < 2
Case 2
Now for x =2,it makes it zero, so not true. Now takes the case of 2<x<3. Lets takes x= 2.5
(2.5−2)(2.5−3)>0
=> −0.25>0
Which is not true so this solution is not good
Case 3
Now lets take the right most part i.e x>3
Lets take x=4
(4−2)(4−3)>0
=> 2>0
So it is good.
Now the solution can either be represented on number line or we can say like this
(−∞,2)∪(3,∞)

Question: Solve y < 2 graphically

Graph of y = 2 is given in the Figure.

Let us select a point, (0, 0) in lower half plane I and putting y = 0 in the given inequality, we see that

1 × 0 < 2 or 0 < 2 which is true.

Thus, the solution region is the shaded region below the line y = 2.

 Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.

Chapter 7

Permutation and Combinations

Fundamental Principle of Counting:

If one experiment has n possible outcomes and another experiment has m possible outcomes, then there are m × n possible outcomes when both of these experiments are performed simultaneously. 

In other words, suppose a job has n parts and the job will be completed only when each part is completed. Further, it is known that the first part can be completed in a₁ ways, the second part can be completed in a₂ ways and so on ...... the nth part can be completed in a_n ways. Then the total number of ways of doing the job is a₁a₂a₃ ... a_n. This is also known as the rule of product.

Note:

• Fundamental Principal of Counting (FPC) is used to calculate the possibilities of an event without actually counting them.
• Basic steps to be used while applying the FPC:

1. Try to identify the independent events involved in a given problem.

2. Find the number of ways of performing or possibilities of occurrence of an event.

3. Multiply these numbers to get the total number of ways of occurrence of these events.

Example 1:

Suppose you have 33 shirts (call them A , B , and C ), and 44 pairs of pants (call them w , x , y , and z ).

Ans.: Then you have

3×4=12

possible outfits:

Aw, Ax, Ay, Az, Bw, Bx, By, Bz, Cw, Cx, Cy, Cz

Example 2:

Suppose you roll a 6 -sided die and draw a card from a deck of 52 cards. There are 6 possible outcomes on the die, and 52 possible outcomes from the deck of cards.

Ans.:

So, there are a total of

6×52=312

possible outcomes of the experiment.

Generalisation of the Addition and the Product Rule

In general, if there are several mutually exclusive events P₁, P₂, P₃, P₄……P_n…etc. with the respective number of ways given as n (P₁), n(P₂), n(P₃), n(P₄)….n(P_n),  then the number of ways in which either P₁ and P₂ and ….. … P_n can occur is given by,

n(E) = n(P₁) + n(P₂) …….. + n(P_n)

Similarly, if there are several mutually independent events P₁, P₂, P₃, P₄……P_n…etc. with the respective number of ways given as n (P₁), n(P₂), n(P₃), n(P₄)….n(P_n), then the number of ways in which  P₁ and P₂ and ….. … P_n  can occur is given by,

n(E) = n(P₁) × n(P₂) …….. × n(P_n)

We must note that all the possible number of ways derived thus, all of them will represent the unique and distinct ways in which the event E will take place.

Event and Its Trials

Suppose we have an event E with ‘m’ possible outcomes. If we conduct this event n number of times, then the number of outcomes of ‘n trials of the event’ will be, mⁿ

Clearly, this is only valid when all the outcomes of the experiment/event E are independent of each other. This would ensure that everytime the event takes place, any of its outcomes are possible.

Question : A coin is tossed 50 times. What is the number of possible outcomes of this experiment?

Answer : A coin toss has two possible outcomes: ‘A heads’ or ‘A tails’. Every toss of the coin is independent of every other toss of the coin, since whenever you toss the coin; there would be two possible outcomes with equal probability. Thus, the no. of possible outcomes of the experiment = 2⁵⁰

Question :

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

(i) repetition of the digits is allowed?

(ii) repetition of the digits is not allowed?

ANSWER:

(i)

There will be as many ways as there are ways of filling 3 vacant places

 in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125

(ii)

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Question :

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

ANSWER:

When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.

Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8

Chapter 8

Binomial Theorem

Binomial Theorem for Positive Integral Indices:

The Binomial Theorem is the method of expanding an Binomial expression that has been raised to any finite power.

Binomial Expression: A binomial expression is an algebraic expression that contains two dissimilar terms. Ex: a + b, a³ + b³,a-b, etc.

The binomial theorem states a formula for the expression of the powers of sums.

From the above representation, we can expand (a + b)ⁿ as given below:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1 b + ⁿC₂ a^n-2 b² + … + ⁿC_n-1 a b^n-1 + ⁿC_n bⁿ

This is the binomial theorem formula for any positive integer n.

 

Some special cases from the binomial theorem can be written as:

• (x + y)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 by+ ⁿC₂ x^n-2 y² + … + ⁿC_n-1 x y^n-1 + ⁿC_n xⁿ
• (x – y)ⁿ = ⁿC₀ xⁿ – ⁿC₁ x^n-1 by + ⁿC₂ x^n-2 y² + … + (-1)^n nC_n xⁿ
• (1 – x)ⁿ = ⁿC₀ – ⁿC₁ x + ⁿC₂ x² – …. (-1)^n nC_n xⁿ

Also, ⁿC₀ = ⁿC_n = 1

However, there will be (n + 1) terms in the expansion of (a + b)ⁿ.

Example: (√2 + 1)⁵ + (√2 − 1)⁵

Sol:

We have

(x + y)⁵ + (x – y)⁵ = 2[5C₀  x⁵ + 5C₂  x³ y²  +  5C₄ xy⁴]

= 2(x⁵ + 10 x³ y² + 5xy⁴)

Now (√2 + 1)⁵ + (√2 − 1)⁵ = 2[(√2)⁵ + 10(√2)³(1)² + 5(√2)(1)⁴]

=58√2

 

Properties:

• The total number of terms in the expansion of (x+y)ⁿ  are (n+1)
• The sum of exponents of x and y is always n.
• ⁿC₀, ⁿC₁, ⁿC₂, … .., ⁿC_n are called binomial coefficients and also represented by C₀, C₁, C₂, ….., C_n
• The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1 , ⁿC₂ = ⁿC_n-2 ,….. etc.

Some other  expansions:

• (x + y)ⁿ + (x−y)ⁿ = 2[C₀ xⁿ + C₂ x^n-1 y² + C₄ x^n-4 y⁴ + …]
• (x + y)ⁿ – (x−y)ⁿ = 2[C₁ x^n-1 y + C₃ x^n-3 y³ + C₅ x^n-5 y⁵ + …]
• (1 + x)ⁿ  = ⁿΣ_r-0 ⁿC_r . x^r = [C₀ + C₁ x + C₂ x² + … C_n x_n]
• (1+x)ⁿ + (1 − x)ⁿ = 2[C₀ + C₂ x²+C₄ x⁴ + …]
• (1+x)ⁿ − (1−x)ⁿ = 2[C₁ x + C₃ x³ + C₅ x⁵ + …]
• The number of terms in the expansion of (x + a)ⁿ + (x−a)ⁿ are (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.
• The number of terms in the expansion of (x + a)ⁿ − (x−a)ⁿ  are (n/2) if “n” is even or (n+1)/2 if “n” is odd.

Question: Find the 4th term in the expansion of (x – 2y)¹².

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿ C _r a^n-r b^r

Here a= x, n =12, r= 3 and b = -2y

By substituting the values we get

T₄ = ¹²C₃ x⁹ (-2y)³

= -1760 x⁹ y³

Chapter 9

Sequences and Series

Sequences and Series:

A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for a_n.

The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by

a₁, a₂, a₃,….,a_n

If a sequence has a finite number of terms then it is known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms.

Series: Let a1, a2, a3,…,an, be a given sequence. Then, the expression

a1 + a2 + a3 +,…+ an + ...

is called the series associated with the given sequence .The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter å (sigma) as means of indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated as

 Sequence relates to the organization of terms in a particular order (i.e. related terms follow each other) and series is the summation of the elements of a sequence.

Examples

Example: Write the first three terms in each of the following sequences defined by

the following:  an = 2n + 5,

Ans.: (i) Here an = 2n + 5

Substituting n = 1, 2, 3, we get

a1 = 2(1) + 5 = 7, a2 = 9, a3 = 11

Therefore, the required terms are 7, 9 and 11.

Example  Let the sequence an be defined as follows:

a1 = 1, an = an – 1 + 2 for n 2.

Find first five terms and write corresponding series.

Solution We have

a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5,

a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9.

Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series

is 1 + 3 + 5 + 7 + 9 +...

Examples

Q.1: If a_n = 2ⁿ, then find the first five terms of the series.

Solution: Given: a_n = 2ⁿ

On substituting n = 1, 2, 3, 4, 5, we get

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

Therefore, the required terms are 2, 4, 8, 16, and 32.

Q.2: Find the sum of odd integers from 1 to 2001.

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_n = n/2 [2a + (n-1)d]

S_n = 1001/2[2×1+(1001-1)2]

S_n = 1001/2[2+1000×2]

S_n = 1001/2[2002]

S_n = 1001 x 1001

S_n = 1002001

Principles of Mathematical induction and simple applications

Question 1:  Prove that m≥ 1 prove that, 1²+ 2² + 3² +4² +……+ m² = m(m + 1)(2m + 1)/6

Solution: Let the given statement be

 P(n) : 1²+ 2² + 3² +4² +……+n² = n(n + 1) (2n + 1) / 6
Let  n= 1 , then P(1) : LHS=1² =1  and RHS=1(1 + 1) (2 × 1 + 1) /6
= 1 × 2 × 3 / 6 = 1

Now LHS=RHS

So, P(1)  is true.

Now, let n=k, and assume X(P) to be true

 i.e.,P(k): 1²+ 2² + 3² +4² +……+k² = k(k+1) (2k+1) / 6                      (1)
We shall now prove that P(k+1) is also true,

so now we have,

Let n=k+1 , then P(k+1) : LHS=

1²+ 2² + 3² +4² +……+k²  + (+1)²
= k(k+1)(2k+1) / 6 + (k+1)²
= k(k+1)(2k+1)+6(k+1)² / 6
= (k+1) {( 2k² + k) + 6(k+1)} /6
= (k+1) (2k² +7k+6)/6
= (k+1) (k+1+1) {2(k+1) +1 }/6
So, P(k + 1) is true, whenever P(k) is true for all natural numbers.

Thus, P(n) is true for all n ÎN

Question 2: Prove that the sum of cubes of n natural numbers is equal to ( n(n+1)²)² for all n natural numbers.

Solution:

Let P(n):

1³+2³+3³+⋯+n³ = (n(n+1)²)²

Step 1: Let n=1 then P(1): LHS=1³=1

RHS=(1(1+1)²)² = 1

So P(1)  is true.

Step 2: Let n=k, then Assume P(k) is true

 i.e.,

P(k): 1³+2³+3³+⋯+k³= (k(k+1)²)² .

Step 3: Let n=k+1, then  P(k+1) : LHS=

                  1³+2³+3³+⋯+k³+(k+1)³ = (k(k+1)²)²+(k+1)³

                 ⇒1³+2³+3³+⋯+k³+(k+1)³=k²(k+1)⁴+(k+1)³

                 = k²(k+1)²+4((k+1)³)⁴

                 =(k+1)²(k²+4(k+1))⁴

                 =(k+1)²(k²+4k+4)⁴

                 = (k+1)²((k+2)²)⁴

                 =(k+1)²(k+1+1)²)⁴

                 =(k+1)²((k+1)+1)²)⁴

                      =RHS

So, P(k + 1) is true, whenever P(k) is true for all natural numbers.

Thus, P(n) is true for all n ÎN

Question3: Show that 1 + 3 + 5 + … + (2n−1) = n²

Solution: 

Step 1: Result is true for n = 1

That is 1 = (1)²  (True)

Step 2: Assume that result is true for n = k

1 + 3 + 5 + … + (2k−1) = k² 

Step 3:  Check for n = k + 1

i.e. 1 + 3 + 5 + … + (2(k+1)−1) = (k+1)² 

We can write the above equation as,

1 + 3 + 5 + … + (2k−1) + (2(k+1)−1) = (k+1)²

Using step 2 result, we get

k² + (2(k+1)−1) = (k+1)²

k² + 2k + 2 −1 = (k+1)² 

k² + 2k + 1 = (k+1)²

(k+1)² = (k+1)² 

L.H.S. and R.H.S. are same.

So, P(k + 1) is true, whenever P(k) is true for all natural numbers.

Thus, P(n) is true for all n ÎN

Question4:  Show that 2²ⁿ-1 is divisible by 3 using the principles of mathematical induction.

To prove: 2²ⁿ-1 is divisible by 3

Assume that the given statement be P(k)

Thus, the statement can be written as P(k) = 2²ⁿ-1 is divisible by 3, for every natural number

Step 1: In step 1, assume n= 1, so that the given statement can be written as

P(1) = 2²⁽¹⁾-1 =  4-1 = 3. So 3 is divisible by 3. (i.e.3/3 = 1)

Step 2: Now, assume that P(n) is true for all the natural number, say k

Hence, the given statement can be written as 

P(k) =  2^2k-1 is divisible by 3.

It means that 2^2k-1 = 3a (where a belongs to natural number)

Now, we need to prove the statement is true for n= k+1

Hence, 

P(k+1) = 2^2(k+1)-1 

P(k+1)= 2^2k+2-1

P(k+1)= 2^2k. 2² – 1

P(k+1)= (2^2k. 4)-1

P(k+1) =3.^2k +(2^2k-1)

The above expression can be written as

P(k+1) =3.2^2k + 3a

Now, take 3 outside, we get

P(k+1) = 3(2^2k + a)= 3b, where “b” belongs to natural number

So, P(k + 1) is true, whenever P(k) is true for all natural numbers.

Thus, P(n) is true for all n ÎN

Question5. n (n + 1) (n + 5) is a multiple of 3

Solution:

We can write the given statement as

P (n): n (n + 1) (n + 5), which is a multiple of 3

If n = 1 we get

1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

Which is true.

Consider P (k) be true for some positive integer k

k (k + 1) (k + 5) is a multiple of 3

k (k + 1) (k + 5) = 3m, where m ∈ N …… (1)

Now let us prove that P (k + 1) is true.

Here

(k + 1) {(k + 1) + 1} {(k + 1) + 5}

We can write it as

= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

So we get

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Substituting equation (1)

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

By multiplication

= 3m + (k + 1) {2k + 10 + k + 2}

On further calculation

= 3m + (k + 1) (3k + 12)

Taking 3 as common

= 3m + 3 (k + 1) (k + 4)

We get

= 3 {m + (k + 1) (k + 4)}

= 3 × q where q = {m + (k + 1) (k + 4)} is some natural number

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

P (k + 1) is true whenever P (k) is true.

Therefore,  P (n) is true for all natural numbers  n.

Question 6. (2n +7) < (n + 3)²

Solution:

We can write the given statement as

P(n): (2n +7) < (n + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)² = 16

Which is true.

Consider P (k) be true for some positive integer k

(2k + 7) < (k + 3)² … (1)

Now let us prove that P (k + 1) is true.

Here

{2 (k + 1) + 7} = (2k + 7) + 2

We can write it as

= {2 (k + 1) + 7}

From equation (1) we get

(2k + 7) + 2 < (k + 3)² + 2

By expanding the terms

2 (k + 1) + 7 < k² + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k² + 6k + 11

Here k² + 6k + 11 < k² + 8k + 16

We can write it as

2 (k + 1) + 7 < (k + 4)²

2 (k + 1) + 7 < {(k + 1) + 3}²

P (k + 1) is true for n=k+1  whenever P (k) is true.

Therefore, P (n) is true for all natural numbers n.

Complex Numbers and Algebra of Complex Numbers

Example 1: Solve the complex number √-4?

Solution: √-4 = √(-1 * 4) = 2√-1 = 2i

Example 2: Solve y²+1 = 0

Solution: y²+1 = 0            

y² = -1

y = √-1 = i

Check:

L.H.S = y²+1 = i² + 1 = -1 + 1 = 0 (R.H.S)

Example 3: Find the value of √-3 + 1.

Solution: √-3 + 1

= √-1 √3 + 1

i√3 + 1

Example 4: Solve x²+16 = 0

Solution:  x²+16 = 0

x²= -16

x =√-16

x = i √16

x = i4

Check: 

L.H.S = x²+16 = (i4)² + 16 = i²4² + 16 = -16 +16 = 0 (R.H.S)

Example 5 : Simplify.

(2i)(−6i)(−7i)(2i)(−6i)(−7i)

Rewrite the expression grouping the real and imaginary parts.

(2i)(−6i)(−7i)=(2)⋅(−6)⋅(−7)⋅i⋅i⋅i

=84⋅i⋅i⋅i(2i)(−6i)(−7i)

=(2)⋅(−6)⋅(−7)⋅i⋅i⋅i=84⋅i⋅i⋅i

Simplify the imaginary part using the property of multiplying powers.

=84i³=84i²i¹=84i3=84i2i1

Recall the definition of i.

Since i²=−1:

=84(−1)i=−84i=84(−1)i=−84i

Example 6:Simplify the principal square root of √−64

Ans:

√−64

Taking the square root and substituting √-1= i

=(√−1). √64=√(−1)⋅ √64=i√64

=8i

The Modulus and the Conjugate of a Complex Numbers

Modulus of a complex number gives the distance of the complex number from the origin in the argand plane, whereas the conjugate of a complex number gives the reflection of the complex number about the real axis in the argand plane.

The product of two complex conjugate numbers is real. = a - ib. , which is real. Thus, the modulus of any complex number is equal to the positive square root of the product of the complex number and its conjugate complex number.

Conjugate of a Complex Number

Conjugate of a complex number z = x + iy is denoted by

= x – iy. It is the reflection of the complex number about the real axis on Argand’s plane or the image of the complex number about the real axis on Argand’s plane. If we replace the ‘i’ with ‘- i’, we get conjugate of the complex number.

Geometrical representation of the complex number is shown in the figure given below:

Properties of conjugate of a complex number:

If z, z₁ and z₂ are complex number, then

Modulus of Complex Number

Modulus of the complex number is the distance of the point on the argand plane representing the complex number z from the origin.

Let P is the point that denotes the complex number z = x + iy.

Then OP = |z| = r = √(x² + y² ).

Properties of Modulus of Complex Number:

(v) For any two complex numbers z₁ and z₂, we have |z₁ + z₂| ≤ |z₁| + |z₂|.

Proof

The distance between the two points z₁ and z₂ in complex plane is | z₁ - z₂ |

If we consider origin, z₁ and z₂ as vertices of a triangle, by the similar argument we have

Geometrical interpretation

Now consider the triangle shown in figure with vertices O, z₁  or z₂ , and z₁ + z₂. We know from geometry that the length of the side of the triangle corresponding to the vector  z₁ + z₂ cannot be greater than the sum of the lengths of the remaining two sides. This is the reason for calling the property as "Triangle Inequality".

It can be generalized by means of mathematical induction to finite number of terms:

 |z₁ + z₂ + z₃ + …. + z_n | ≤ |z₁| + |z₂| + |z₃| + … + |z_n| for n = 2,3,….

Note:

1. |z| > 0.

2. All the complex number with same modulus lie on the circle with centre origin and radius r = |z|.

Application:

Spectrum Analyzer

Those cool displays you see when music is playing? Yes, Complex Numbers are used to calculate them! Using something called "Fourier Transforms".

In fact many clever things can be done with sound using Complex Numbers, like filtering out sounds, hearing whispers in a crowd and so on.

It is part of a subject called "Signal Processing".

Example 1: Find the conjugate of the complex number z = (1 + 2i)/(1 – 2i).

Solution: z = (1 + 2i)/(1 – 2i)

Rationalising given complex number we have

⇒ z = ((1 + 2i)/(1 – 2i) )× (1 + 2i)/(1 + 2i)

⇒ z = (1 + 2i)²/(1² – (2i)²)

⇒ z = (1 + 4i² + 4i)/(1 + 4)

⇒ z = (1 – 4 + 4i)/(1 + 4)

⇒ z = (-3 + 4i)/5

⇒z= (- 3 – 4i)/5.

Example 2: Find the modulus of the complex number z = (3 – 2i)/2i

Solution: z = (3 – 2i)/2i

⇒ z = (3 )/2i – 2i/2i

⇒ z = 3/2i – 1

⇒ z = 3i/(2i² ) – 1

⇒ z = (-3i/2) – 1

Example 3: If z + |z| = 1 + 4i, then find the value of |z|.

Solution: Let z = x + iy

⇒ z + |z| = 1 + 4i

⇒ x + iy + |x + iy| = 1 + 4i

⇒ x + iy + √(x² + y² ) = 1 + 4i

⇒ y = 4 and x + √(x² + y² ) = 1

⇒ x + √(x² + 4² ) = 1

⇒ √(x² + 4² ) = 1 – x

Squaring both side we have

⇒ x² + 4² = 1 + x² – 2x

⇒ 2x = -15

or x = -15/2

Argand Plane and Polar Representation

The plane having a complex number assigned to each of its point is called the

complex plane or the Argand plane.

Argument of Complex Numbers 

The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. It is denoted by “θ” or “φ”. It is measured in the standard unit called “radians”.

The real and complex components of coordinates are found in terms of r and θ where r is the length of the vector, and θ is the angle made with the real axis.

The point P is uniquely determined by the ordered pair of real numbers (r, ), called the polar

coordinates of the point P.

Polar Form of Complex Numbers 

The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. It is denoted by “θ” or “φ”. It is measured in the standard unit called “radians”.

The real and complex components of coordinates are found in terms of r and θ where r is the length of the vector, and θ is the angle made with the real axis.

We have, x = r cos q, y = r sin q and therefore, z = r (cos q + i sin q).

The latter is said to be the polar form of the complex number.

q is called the argument (or amplitude) of z which is denoted by arg z. and

r = |z| =√(x² + y²) is the modulus of z

For any complex number z ¹ 0, there corresponds only one value of q in 0 £ q < 2p.

The value of q such that – p < q £ p, called principal argument of z and is denoted by arg z.

Properties of Argument of Complex Numbers

Suppose that z be a nonzero complex number and n be some integer, then

Let us assume, z₁ and z₂ be the two complex numbers, the following identities are:

Solving Quadratic equations in complex numbers

We can solve quadratic equations with complex coefficients using the quadratic formula. If the discriminant is zero, the equation has one repeated root. If all the coefficients are real, the root will be real. A general quadratic with complex coefficients can have any combination of real and nonreal roots.

The Quadratic Formula is   The expression under the radical, is called the discriminant. You can use the discriminant to determine the number and type of solutions an equation has.

Using the discriminant, we identify the three different cases of quadratic equations as follows.

Solving Equations with Different Types of Solutions

4. Nonreal Quadratic Equations with Real Negative Discriminants

 

Square Root and Cube Root Of Complex Numbers:

Square roots of a complex number

Let the square root of a + ib be x + iy

That is = √[a+ib] = x + iy where x, y ∈ R

a + ib =  ( x + iy )²  = x²  - y²  + i2xy

Equating real and imaginary parts, we get 

x² - y² = a and 2xy = b

(x²  + y² )² =  (x² - y² ) ²  + 4x² y²  = a²  + b²

x²  + y² =  √[a²+b²], since x² + y² is positive 

Solving x² + y² = a  and  x²+ y² =√[a²+b²], we get

Since 2xy = b  it is clear that both x and y will have the same sign when b is positive, and x and y have different signs when b is negative.

Formula for finding square root of a complex number

Note

Example 1.

Find the square root of 6 - 8i.

Solution

We compute |6 - 8i| = √[6² + (-8)²] = 10

and applying the formula for square root, we get

Example 2: Determine the square root of complex number 3 + 4i.

Solution: To determine the square root of 3 + 4i, we will determine its magnitude and compare with a + ib.

|3 + 4i| = √(3² + 4²) = √(9 + 16) = √25 = 5; a = 3 and b = 4 > 0

Therefore, using formula for square root of complex numbers, we have

√(3 + 4i) = ± (√[√(5 + 3)/2] + i√[√(5 - 3)/2] )

= ± (√(8/2) + i √(2/2))

= ±(√4 + i√1)

= ± (2 + i)

Answer: Hence, √(3 + 4i) = ± (2 + i)

Cube Root of Unity Values:

Suppose the cube root of 1 is “x”,

i .e  .,  ³√1 = x.

According to the general cube roots definition,

                            x³ = 1

• •  x³ – 1 = 0

since (a³ – b³) = (a – b) ( a² + ab + b²)

Now, (x³ – 1³) = 0

• • (x – 1)( x² + x + 1) = 0

Therefore,

 x = 1  Or   ( x² + x + 1) = 0

• • x=1  or  x = [(-1) ± √(1²-4.1.1)]/2
  • x=1 or x= [-1 ± √-3]/2
  • x=1 or  x= -1/2 ± i√(3)/2

Therefore, the three cube roots of unity are:

1,  ω= -1/2+i√(3)/2,  ω²= -1/2 – i√(3)/2

Properties of Cube roots of unity

1) One imaginary cube roots of unity is the square of the other imaginary cube root.

= ¼ [(-1)² – 2 × 1 × √3 i + ( √3 i)²] = ¼(1 – 2√ 3i – 3) = (-1-√ 3 i) /2

And

= ¼ [(-1)² + 2 × 1 × √3 i + ( √3 i)²] = ¼(1 + 2√ 3i – 3) = (-1+ √ 3 i) /2

i.e., ω²= (ω²)²

2) If two imaginary cube roots are multiplied then the product we get is equal to 1.

Let’s take ω = (-1 + √3 i ) /2

ω² = (-1 – √3 i ) /2

Now we will get the product of two imaginary cube roots as ω x ω² = [(-1 + √3 i ) / 2]x [(-1 – √3 i ) /2]

Or ω³ = ¼[(-1)² – (√3 i)²]

= ¼[( 1 – 3i²) = ¼ x 4 = 1.

3) As there are three cube roots of unity, their sum is zero, let’s see how

= 1 + [(-1 + √3 i ) /2] + [(-1 – √3 i ) /2]

Or

1 + ω + ω² = 1 – ½ + (√3 i ) /2 – ½ – (√3 i ) /2

= 0

4) conjugate of One imaginary cube roots of unity is equal to the other imaginary cube root.

Euler’s form of a Complex Number

e^iƟ = cos Ɵ + i sin Ɵ

This form makes the study of complex numbers and its properties simple. Any complex number can be expressed as

z = x + iy (Cartesian form)

= |z| [cos Ɵ + I sin Ɵ] (polar form)

= |z| e^iƟ

De Moivre’s Theorem and its Applications

(a) De Moivre’s Theorem for integral index. If n is a integer, then (cos Ɵ + i sin Ɵ)ⁿ = cos (nƟ) + i sin (nƟ)

(b) De Moivre’s Theorem for rational index. If n is a rational number, then the value of or one of the values of

(cosƟ + isinƟ)ⁿ is cos (nƟ) + i sin (nƟ).

In fact, if n = p/q where p, q ϵ I, q > 0 and p,q have no factors in common, then (cos Ɵ + i sin Ɵ)ⁿ has q distinct values, one of which is cos (nƟ) + i sin (nƟ)

The nth Roots of Unity

By an nth root of unity we mean any complex number z which satisfies the equation zⁿ = 1 (1)

Since an equation of degree n has n roots, there are n values of z which satisfy the equation (1). To obtain these n values of z, we write 1 = cos (2kπ) + I sin (2kπ)

Where k ϵ I and

[using the De Moivre’s Theorem]

Where k = 0, 1, 2, …., n -1.

Note

The values of (cos Ɵ + I sin Ɵ)^p/q where p, q ϵ I, q ≠ 0, hcf (p,q) = 1 are given by

Where k = 0, 1, 2, ….., q -1.

Graphical methods and Solution of Linear Inequalities in Two Variables

Graph of Linear inequalities in Two Variable

A linear equation in two variable is of the form

ax+by+c=0

We have already studied in Coordinate geometry that this can be represented by a straight line in xy- plane. All the points on the straight line are the solutions of this linear equation.

we can similarly find the solution set graphically for the linear inequalities in the below form

ax+by+c<0

ax+by+c>0

ax+by+c≥0

ax+by+c≤0

How to find the solution graphically for Linear inequalities in Two Variable.:

1. Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign.
   ax+by+c=0
    
   • This can be done easily by Point on the x-axis( x,0) and point on the y axis ( 0,y)
   • Point on x-axis given by ax+b(0)+c=0 or x=−c/a or (−c/a,0)
   • Point on y-axis given by a(0)+by+c=0 or y=−c/b or (0,−c/b)
   • Locate these point on cartesian plane and join them to find the line
2. Use a dashed or dotted line if the problem involves a strict inequality, < or >.
3. Otherwise, use a solid line to indicate that the line   itself constitutes part of the solution.
4. Pick a  point lying in one of the half-planes determined by the line sketched in step 1 and substitute the values of x and y into the given inequality.
   Use the origin whenever possible.
5. If the inequality is satisfied, the graph of the inequality includes the half-plane containing the test point.
   Otherwise, the solution includes the half-plane not containing the test point

Example
Determine the solution set for the inequality
x+y>1
Solution
1) Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign.
i.e x+y=1
 

 2) Pick the test point as origin (0,0), and put into the inequality
0+0>1
0>1
Which is false
So the solution set is other half plane of the line

How to find the solution graphically for pair of  Linear inequalities in Two Variable.

ax+by+c<0
px+qy+c<0
The solution set of a system of linear inequalities in two variables x and y is the set of all points (x, y) that satisfy each inequality of the system.
Step

1. Find the graphical solution for each inequality independently using the technique described above
2. Now determine the region in common with each solution set

Example
Find the solution of the below system of inequalities
2x+3y>1
x+2y>2
x>1
Solution
1) for 2x+3y>1, Solving using the above method solution is

2) for x+2y>2, Solving using the above method solution is

3) for x>1, Solving using the above method solution is

4) Now we draw these on the single graph and can determine the common region

Example:

Solve the following system of inequalities

8x + 3y £ 100 ......... (1)

           x ³ 0 ......... (2)

           y ³ 0 ... …. (3)

Ans:

Let 8x + 3y = 100

The inequality 8x + 3y £ 100 represents the shaded region below the line, including the

points on the line 8x +3y =100.

Since x ³ 0, y ³ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities.

 

Question: Solve the following system of inequalities graphically

x + 2y £ 8 ... (1)

2x + y £ 8 ... (2)

x > 0 ... (3)

y > 0 ... (4)

Solution:

We draw the graphs of the lines

x + 2y = 8  

and

 2x + y = 8.

The inequality (1) and (2) represent the region below the two lines, including the point on the respective lines.

Since x ³ 0, y ³ 0, every point in the shaded region in the first quadrant represent

a solution of the given system of inequalities.

Note:

1 The region containing all the solutions of an inequality is called thesolution region.

2. In order to identify the half plane represented by an inequality, it is just sufficient to take any point (a, b) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point (0, 0) is preferred.

3. If an inequality is of the type ax + by ³ c or ax + by £ c, then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region.

4. If an inequality is of the form ax + by > c or ax + by < c, then the points on the line ax + by = c are not to be included in the solution region. So draw a broken ordotted line in the solution region.

Example: Solve – 8 £ 5x – 3 < 7.

Solution : In this case, we have two inequalities,

 – 8 £ 5x – 3 and 5x – 3 < 7,

which we will solve simultaneously.

We have – 8 £ 5x –3 < 7

or –5 £ 5x < 10 or –1 £ x < 2

Example:

A manufacturer has 600 litres of a 12% solution of acid. How many litres

of a 30% acid solution must be added to it so that acid content in the resulting mixture

will be more than 15% but less than 18%?

Solution: Let x litres of 30% acid solution is required to be added. Then

Total mixture = (x + 600) litres