4. Sum to n Terms of Special Series

Sum to n Terms of Special Series

sum of first n terms of some special series,

(i) 1 + 2 + 3 +… + n (sum of first n natural numbers)

(ii) 1² + 2² + 3² +… + n² (sum of squares of the first n natural numbers)

(iii) 1³ + 2³  + 3³  +… + n³ (sum of cubes of the first n natural numbers).

Proof:

(i) 1 + 2 + 3 +… + n (sum of first n natural numbers)

Sum of First n Natural Numbers

Natural numbers are: 1, 2, 3, 4,….

Sum of these natural numbers can be written as: 1 + 2 + 3 + 4 +….

This is an AP with first term 1 and common difference 1.

i.e. a = 1 and d = 2 – 1 = 1

Sum of first n terms of an AP = n/2 [2a + (n – 1)d]

Now,

S_n = 1 + 2 + 3 + 4 + ….. + n

S_n = n/2 [2a + (n – 1)d]

Substituting a = 1 and d = 1,

S_n = (n/2) [2(1) + (n – 1)(1)]

= (n/2) [2 + n – 1]

= n(n + 1)/2

Therefore, the sum of first n natural numbers = n(n + 1)/2

Sum of Squares of The First n Natural Numbers

(ii) 1² + 2² + 3² +… + n² (sum of squares of the first n natural numbers)

The squares of natural numbers are: 1², 2², 3², 4²,…

Or

1, 4, 9, 16, ….

We can express the sum of n terms as: 1² + 2² + 3² +…+ n²

This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.

Let’s find the sum of this series by considering an expression given below:

k³ – (k – 1)³ = 3k² – 3k + 1

Substituting k = 1,

1³ – (1 – 1)³ = 3(1)² – 3(1) + 1

1³ – 0³ = 3(1)² – 3(1) + 1….(i)

Substituting k = 2,

2³ – (2 – 1)³ = 3(2)² – 3(2) + 1

2³ – 1³ = 3(2)² – 3(2) + 1….(ii)

Substituting k = 3,

3³ – (3 – 1)³ = 3(3)² – 3(3) + 1

3³ – 2³ = 3(3)² – 3(3) + 1….(iii)

Substituting k = 4,

4³ – (4 – 1)³ = 3(4)² – 3(4) + 1

4³ – 3³ = 3(4)² – 3(4) + 1….(iv)

Substituting k = n,

n³ – (n – 1)³ = 3(n)² – 3(n) + 1

Now, adding both sides of these equations together, we get;

1³ – 0³ + 2³ – 1³ + 3³ – 2³ + … + n³ – (n – 1)³ = 3(1² + 2² + 3² + 4² + … + n²) – 3(1 + 2 + 3 + 4 + … + n) + n(1)

n³ – 0³ = 3(1² + 2²+ 3² + 4² + … + n²) – 3(1 + 2 + 3 + 4 + … + n) + n

Here,

represents the sum of first n natural numbers and is equal to n(n + 1)/2.

So,

Rearranging the terms,

= (1/6) (2n³ + 3n² + n)

= (1/6) [n(2n² + 3n + 1)]

= (1/6)[n(n + 1)(2n + 1)]

Therefore, the sum of squares of first n natural numbers = [n(n + 1)(2n + 1)]/6

Sum of Cubes of The First n Natural Numbers

(iii) 1³ + 2³  + 3³  +… + n³ (sum of cubes of the first n natural numbers).

The squares of natural numbers are: 1³, 2³, 3³, 4³,…

Or

1, 8, 27, 64,….

We can express the sum of n terms as: 1³ + 2³ + 3³ +…+ n³

This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.

Let’s find the sum of this series by considering an expression given below:

(k + 1)⁴ – k⁴ = 4k³ + 6k² + 4k + 1

Substituting k = 1, 2, 3, …, n

2⁴ – 1⁴ = 4(1)³ + 6(1)² + 4(1) + 1

3⁴ – 2⁴ = 4(2)³ + 6(2)² + 4(2) + 1

4⁴ – 3⁴ = 4(3)³ + 6(3)² + 4(3) + 1

(n – 1)⁴ – (n – 2)⁴ = 4(n – 2)³ + 6(n – 2)² + 4(n – 2) + 1

n⁴ – (n – 1)⁴ = 4(n – 1)³ + 6(n – 1)² + 4(n – 1) + 1

(n + 1)⁴ – n⁴ = 4n³ + 6n² + 4n + 1

By adding both sides of these equations, we get;

2⁴ – 1⁴ + 3⁴ – 2⁴ + 4⁴ – 3⁴ + …. + (n + 1)⁴ – n⁴ = 4(1)³ + 6(1)² + 4(1) + 1 + 4(2)³ + 6(2)² + 4(2) + 1 + 4(3)³ + 6(3)² + 4(3) + 1 + …. + 4n³ + 6n² + 4n + 1

(n + 1)⁴ – 1⁴ = 4(1³ + 2³ + 3³ +…+ n³) + 6(1² + 2² + 3² + …+ n²) + 4(1 + 2 + 3 +…+ n) + n

We know that,

And

Thus,

By rearranging the terms,

= (1/4) [n⁴ + 4n³ + 6n² + 4n – 2n³ – 3n² – n – 2n² – 2n – n]

= (1/4) [n⁴ + 2n³ + n²]

= (1/4)[n²(n² + 2n + 1)]

= (1/4)[n²(n + 1)²]

Therefore, the sum of cubes of first n natural numbers = [n(n + 1)]²/4

Question:

Find the sum to n terms of the series: 2 + 5 + 14 + 41 +….

Solution:

2 + 5 + 14 + 41 +….

The difference between two consecutive terms of this series is: 3, 9, 27, ….

Let S_n be the sum of its n terms and an be its nth term. Then,

S_n =2 + 5 + 14 + 41 + … + a_n….(i)

And

S_n = 2 + 5+ 14 + 41 + … + a_{n – 1} + a_n….(ii)

Subtracting equation (ii) from (i), we get

0 = 2 + [3 + 9 + 27 + … + (n – 1) term] – a_n

⇒ a_n = 2 + [3 + 9 + 27 +… + (n- 1) term]

Here, 3 + 9 + 27 + … is a geometric series.

So, a_n = 2 + [3(3^n-1 – 1)/2]

= [4 + 3ⁿ – 3]/2

= (1 + 3ⁿ)/2

Now, we need to find the sum of the series whose general term is (1 + 3ⁿ)/2

Sn = [(1 + 3)/2] + [(1 + 3²)/2] + [(1 + 3³)/2] + …. + (1 + 3ⁿ)/2

= (1/2) [(3 + 3² + 3³ + …. + 3ⁿ) + (1 + 1 + 1 + … + n)]

= (1/2) {[3(3ⁿ – 1)/(3 – 1)] + n}

= (1/2) [(3/2)(3ⁿ – 1) + n]

= [(3ⁿ⁺¹ – 3 + 2n)/4]

Therefore, the sum of the given series = (3ⁿ⁺¹ + 2n – 3)/4