1. Binomial Theorem for Positive Integral Indices

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 ExpressionA 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 an-1 b + nC2 an-2 b2 + … + nCn-1 a bn-1 + nCn 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 xn-1 by+ nC2 xn-2 y2 + … + nCn-1 x yn-1 + nCn xn
  • (x – y)n = nC0 xn – nC1 xn-1 by + nC2 xn-2 y2 + … + (-1)n nCn xn
  • (1 – x)n = nC0 – nC1 x + nC2 x2 – …. (-1)n nCn xn

Also, nC0 = nCn = 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(x+ 10 x3 y+ 5xy4)

Now (√2 + 1)+ (√2 − 1)= 2[(√2)+ 10(√2)3(1)+ 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, … .., nCn are called binomial coefficients and also represented by C0, C1, C2, ….., Cn
  • The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC= nCn, nC= nCn-1 , nC= nCn-2 ,….. etc.

Some other  expansions:

  • (x + y)+ (x−y)= 2[C0 x+ C2 xn-1 y+ C4 xn-4 y+ …]
  • (x + y)– (x−y)= 2[C1 xn-1 y + C3 xn-3 y+ C5 xn-5 y+ …]
  • (1 + x)n  nΣr-0 nC. x= [C+ C1 x + C2 x+ … Cn xn]
  • (1+x)+ (1 − x)= 2[C0 + C2 x2+C4 x+ …]
  • (1+x)− (1−x)= 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 Tr+1 in the binomial expansion is given by Tr+1 = r an-r br

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

By substituting the values we get

T4 = 12C3 x9 (-2y)3

= -1760 x9 y3

2. Pascal’s triangle

Pascal’s triangle

An expression consisting of two terms, connected by + or – sign is called binomial expression. Binomial Theorem. If a and b are real numbers and n is a positive integer, then. The general term of (r + 1)th term in the expression is given by. Tr+1 = nCr an-r br.

Example: (a + b)7 = 7C0 a7+ 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6ab6 + 7C7b7

3. General and Middle Terms

General and Middle Terms

General Term in binomial expansion:

We have (x + y)= nC0 x+ nC1 xn-1 . y + nC2 xn-2 . y+ … + nCn yn

General Term = Tr+1 = nCr xn-r . yr

  • General Term in (1 + x)n is nCxr
  • In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .

Example: Find the number of terms in (1 + 2x +x2)50

Sol:

(1 + 2x + x2)50 = [(1 + x)2]50 = (1 + x)100

The number of terms = (100 + 1) = 101

ExampleFind the fourth term from the end in the expansion of (2x – 1/x2)10

Sol:

Required term =T10 – 4 + 2 = T8 = 10C(2x)(−1/x2)= −960x-11

 

Middle Term(S) in the expansion of (x+y) n

  • If n is even then (n/2 + 1) Term is the middle Term.
  • If n is odd then [(n+1)/2]th and [(n+3)/2)th terms are the middle terms.

i.e., (n+12)thterm and (n+12+1)thterm

ExampleFind the middle term of (1 −3x + 3x– x3)2n

Sol:

(1 − 3x + 3x– x3)2n = [(1 − x)3]2n = (1 − x)6n

Middle Term = [(6n/2) + 1] term = 6nC3n (−x)3n

Determining a Particular Term:

  • In the expansion of(axp + b/xq)the coefficient of xm is the coefficient of Tr+1 where r = [(np−m)/(p+q)]
  • In the expansion of (x + a)n, Tr+1/Tr = (n – r + 1)/r . a/x

Independent Term

The term Independent of in the expansion of [ax+ (b/xq)]n is

Tr+1 = nCr an-r br, where r = (np/p+q) (integer)

Example: Find the independent term of x in (x+1/x)6

Sol:

r = [6(1)/1+1] = 3

The independent term is 6C= 20

Example :In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.

Solution:

We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

Tr+1 = m+n Cr 1m+n-r ar

m+n Cr ar…………. (i)

Now we have that the general term for the expression is,

Tr+1 =  m+n Cr ar

Now, For coefficient of am

Tm+1 =  m+n Cm am

Hence, for coefficient of am, value of r = m

So, the coefficient is m+n m

Similarly, Coefficient of an is m+n n

Example: Write the general term in the expansion of (x2 – y x)12, x ≠ 0.

Solution:

The general term Tr+1 in the binomial expansion is given by Tr+1 = r an-r br

Here n = 12, a= x2 and b = -y x

Substituting the values we get

Tn+1 =12Cr × x2(12-r) (-1)r yr xr

= -1r 12c.x24 –2r. yr

Example: Which is larger (1.01)1000000 or 10,000?

Ans.:

Numerically greatest term in the expansion of (1+x)n:

  • If [(n+1)|x|]/[|x|+1] = P, is a positive integer then Pth term and (P+1)th terms are numerically the greatest terms in the expansion of (1+x)n
  • If[(n+1)|x|]/[|x|+1] = P + F, where P is a positive integer and 0 < F < 1 then (P+1)th term is numerically the greatest term in the expansion of (1+x)n.

Example: Find the numerically greatest term in (1-3x)10 when x = (1/2)

Sol:

[(n + 1)|α|] / [|α| + 1] = (11 × 3/2)/(3/2+1) = 33/5 = 6.6

Therefore, T7 is the numerically greatest term.

T6 + 1 = 10C. (−3x)= 10C6 . (3/2)6

Ratio of Consecutive Terms/Coefficients:

Coefficients of xr and xr + 1 are nCr – 1 and nCr respectively.

(nCr / nCr – 1) = (n – r + 1) / r

Example: If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:7:42 then find the value of n.

Sol:

Let (r – 1)th, (r)th and (r + 1)th be the three consecutive terms.

Then, the given ratio is 1:7:42

Now (nCr-2 / nCr – 1) = (1/7)

(nCr-2 / nCr – 1) = (1/7) [(r – 1)/(n − r+2)] = (1/7) n−8r+9=0 → (1)

And,

(nCr-1 / nCr) = (7/42) [(r)/(n – r +1)] =(1/6) n−7r +1=0 → (2)

From (1) & (2), n = 55

Applications of Binomial Theorem

Binomial theorem has a wide range of applications in Mathematics like finding the remainder, finding digits of a number, etc. The most common binomial theorem applications are:

Finding Remainder using Binomial Theorem

Example: Find the remainder when 7103 is divided by 25

Sol:

(7103 / 25) = [7(49)51 / 25)] = [7(50 − 1)51 / 25]

= [7(25K − 1) / 25] = [(175K – 25 + 25−7) / 25]

= [(25(7K − 1) + 18) / 25]

The remainder = 18.

Example: If the fractional part of the number (2403 / 15) is (K/15), then find K.

Sol:

(2403 / 15) = [2(24)100 / 15]

= 8/15 (15 + 1)100 = 8/ 15 (15λ + 1) = 8λ + 8/15

8λ is an integer, fractional part = 8/15

So, K = 8.

Finding Digits of a Number

Example: Find the last two digits of the number (13)10

Sol:

(13)10 = (169)= (170 − 1)5

= 5C0 (170)− 5C(170)+ 5C(170)− 5C(170)+ 5C(170) − 5C5

= 5C(170)− 5C(170)+ 5C(170)− 5C(170)+ 5(170) − 1

A multiple of 100 + 5(170) – 1 = 100K + 849

The last two digits are 49.

Relation Between two Numbers

Example: Find the larger of 9950 + 10050 and 10150

Sol:

10150 = (100 + 1)50 = 10050 + 50 . 10049 + 25 . 49 . 10048 + …

9950 = (100 − 1)50 = 10050 – 50 . 10049 + 25 . 49 . 10048 − ….

10150 – 9950 = 2[50 . 10049 + 25(49) (16) 10047 + …]

= 10050 + 50 . 49 . 16 . 10047 + … >10050

10150 – 9950 > 10050

10150 > 10050 + 9950

Divisibility Test

Example: Show that 11+ 911 is divisible by 10.

Sol:

11+ 911 = (10 + 1)+ (10 − 1)11

= (9C. 10+ 9C. 10+ … 9C9) + (11C. 1011 − 11C. 1010 + … −11C11)

= 9C. 10+ 9C. 10+ … + 9C. 10 + 1 + 1011 − 11C. 1010 + … + 11C10 . 10−1

= 10[9C. 10+ 9C. 10+ … + 9C+ 11C. 1010 − 11C. 10+ … + 11C10]

= 10K, which is divisible by 10.

Formulae:

  • The number of terms in the expansion of (x+ x+ … xr)n is (n + r − 1)Cr – 1
  • Sum of the coefficients of (ax + by)n is (a + b)n

If f(x) = (a+ a1x + a2x+ …. + amxm)n then

  • (a) Sum of coefficients = f(1)
  • (b) Sum of coefficients of even powers of x is: [f(1) + f(−1)] / 2
  • (c) Sum of coefficients of odd powers of x is [f(1) − f(−1)] / 2

Multinomial Theorem

Using binomial theorem, we have

(x + a)n

=nr = 0nCr xn – r ar, nN

nr = 0 [n! / (n − r)!r!] xn – rar

nr + s =n [n! / r!s!] xs ar, where s = n – r.

This result can be generalized in the following form:

(x1 + x2 + … +xk)n

= ∑r1 + r2 + …. + rk = n [n! / r1!r2!…rk!] x1r1 x2r2 …xkrk

The general term in the above expansion is

[(n!) / (r1! r2! r3! … rk!)] x1r1 x2r2 x3r3… xkrk

The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation.

r1+r+ … + r= n, because each solution of this equation gives a term in the above expansion. The number of such solutions is n + k – 1Ck −1.

PARTICULAR CASES

Case-1:

 

The above expansion has n+3-1C3-1 = n + 2C2 terms.

Case-2: 

There are n + 4 – 1C4 – 1 = n + 3C3 terms in the above expansion.

REMARK: The greatest coefficient in the expansion of (x+ x+ … + xm)n is [(n!) / (q!)m – r{(q+1)!}r], where q and r are the quotient and remainder, respectively when n is divided by m.

Multinomial Expansions

Consider the expansion of (x + y + z)10. In the expansion, each term has different powers of x, y, and z and the sum of these powers is always 10.

One of the terms is λx2y3z5. Now, the coefficient of this term is equal to the number of ways 2x′s, 3y′s, and 5z′s are arranged, i.e., 10! (2! 3! 5!). Thus,

(x+y+z)10 = ∑(10!) /  (P1! P2! P3!) xP1 yP2 zP3

Where P1 + P2 + P3 = 10 and 0 ≤ P1, P2, P3 ≥ 10

In general,

(x+ x+ … xr)= ∑ (n!) / (P1! P2! … Pr!) xP1 xP2 … xPr

Where P1 + P2 + P3 + … + Pr = n and 0 ≤ P1, P2, … Pr ≥ n

Number of Terms in the Expansion of (x1 + x2 + … + xr)n

From the general term of the above expansion, we can conclude that the number of terms is equal to the number of ways different powers can be distributed to x1, x2, x…., xn such that the sum of powers is always “n”.

Number of non-negative integral solutions of x+ x+ … + x= n is n +r – 1Cr – 1.

For example, number of terms in the expansion of (x + y + z)3 is 3 + 3 -1C3 – 1 =  5C= 10

As in the expansion, we have terms such as

As xy0 z0, x0 y1 z2, x0 y2 z1, x0 y3 z0, x1 y0 z2, x1 y1 z1, x1 y2 z0, x2 y0 z1, x2 y1 z0, x3 y0 z0.

Number of terms in (x + y + z)n is n + 3 – 1C3 – 1 n + 2C2.

Number of terms in (x + y + z + w)n is n + 4 – 1C4 – 1 = n + 3C3 and so on.

Example. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

Solution:

In order to show that 9n+1 – 8n – 9 is divisible by 64, it has to be show that 9n+1 – 8n – 9 = 64 k, where k is some natural number

Using binomial theorem,

(1 + a)m = mC0 + mC1 a + mC2 a2 + …. + m am

For a = 8 and m = n + 1 we get

(1 + 8)n+1 = n+1C0 + n+1C1 (8) + n+1C2 (8)2 + …. + n+1 n+1 (8)n+1

9n+1 = 1 + (n + 1) 8 + 82 [n+1C2 + n+1C3 (8) + …. + n+1 n+1 (8)n-1]

9n+1 = 9 + 8n + 64 [n+1C2 + n+1C3 (8) + …. + n+1 n+1 (8)n-1]

9n+1 – 8n – 9 = 64 k

Where k = [n+1C2 + n+1C3 (8) + …. + n+1 n+1 (8)n-1] is a natural number

Thus, 9n+1 – 8n – 9 is divisible by 64, whenever n is positive integer.

Hence the proof

Example. Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.

Solution:

We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3

Putting a = 3x2 & b = -a (2x-3a), we get

[3x2 + (-a (2x-3a))]3

= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2 + (-a (2x-3a))3

= 27x6 – 27ax(2x-3a) + 9a2x(2x-3a)2 – a3(2x-3a)3

= 27x6 – 54ax5 + 81a2x4 + 9a2x(4x2-12ax+9a2) – a[(2x)3 – (3a)3 – 3(2x)2(3a) + 3(2x)(3a)2]

= 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108a3x3 + 81a4x2 – 8a3x3 + 27a6 + 36a4x2 – 54a5x

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6

Thus, (3x2 – 2ax + 3a2)3

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6