# Sum to n Terms of Special Series

In mathematics, we may come across different types of series such as arithmetic series, geometric series, harmonic series, etc. Apart from these, we can observe some special series for which we can find the sum of the terms using different techniques. In this article, you will learn three most commonly used special series and derivation of formulas to find the sum of these series upto n terms along with the solved example.

## Sum of n Terms of Special Series

Some special series are given below:

(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)

Let us find the sum upto n terms of special series mentioned here, one by one.

### 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 +â€¦.

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,

Sn = 1 + 2 + 3 + 4 + â€¦.. + n

Sn = n/2 [2a + (n – 1)d]

Substituting a = 1 and d = 1,

Sn = (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

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

Or

1, 4, 9, 16, â€¦.

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.

Letâ€™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

$n^{3}=3\sum_{k=1}^{n}k^{2}-3\sum_{k=1}^{n}k+n$

Here, $\sum_{k=1}^{n}k$ represents the sum of first n natural numbers and is equal to n(n + 1)/2.

So,

$n^{3}=3\sum_{k=1}^{n}k^{2}-3[\frac{n(n+1)}{2}]+n$

Rearranging the terms,

$\sum_{k=1}^{n}k^{2}=\frac{1}{3}[n^{3}+3[\frac{n(n+1)}{2}]-n]$ $\sum_{k=1}^{n}k^{2}=\frac{1}{6}[2n^{3}+3n^2+3n-2n]$

= (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

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.

Letâ€™s find the sum of this series by considering an expression given below:

(k + 1)4 â€“ k4 = 4k3 + 6k2 + 4k + 1

Substituting k = 1, 2, 3, …, n

24 â€“ 14 = 4(1)3 + 6(1)2 + 4(1) + 1

34 â€“ 24 = 4(2)3 + 6(2)2 + 4(2) + 1

44 â€“ 34 = 4(3)3 + 6(3)2 + 4(3) + 1

…..

(n â€“ 1)4 â€“ (n â€“ 2)4 = 4(n â€“ 2)3 + 6(n â€“ 2)2 + 4(n â€“ 2) + 1

n4 â€“ (n â€“ 1)4 = 4(n â€“ 1)3 + 6(n â€“ 1)2 + 4(n â€“ 1) + 1

(n + 1)4 â€“ n4 = 4n3 + 6n2 + 4n + 1

By adding both sides of these equations, we get;

24 â€“ 14 + 34 â€“ 24 + 44 â€“ 34 + …. + (n + 1)4 â€“ 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

$n^4 + 4n^3 + 6n^2 + 4n + 1 – 1 =4\sum_{k=1}^{n}k^3 + 6\sum_{k=1}^{n}k^2 + 4\sum_{k=1}^{n}k+n$

We know that,

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

And

$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$

Thus,

$n^4 + 4n^3 + 6n^2 + 4n = 4\sum_{k=1}^{n}k^3 + 6[\frac{n(n+1)(2n+1)}{6}]+4[\frac{n(n+1)}{2}]+n$

By rearranging the terms,

$4\sum_{k=1}^{n}k^3 = n^4 + 4n^3 + 6n^2 + 4n – 6[\frac{n(n+1)(2n+1)}{6}]-4[\frac{n(n+1)}{2}]-n\\\sum_{k=1}^{n}k^3 = \frac{1}{4}[n^4 + 4n^3 + 6n^2 + 4n – n (2n^2 + 3n + 1) – 2n(n + 1) – n]$

= (1/4) [n4 + 4n3 + 6n2 + 4n â€“ 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

Also, get the Sum of cubes formula here.

### Solved Example

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, â€¦.

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

Sn =2 + 5 + 14 + 41 + â€¦ + an….(i)

And

Sn = 2 + 5+ 14 + 41 + â€¦ + an â€“ 1 + an….(ii)

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

0 = 2 + [3 + 9 + 27 + â€¦ + (n â€“ 1) term] â€“ an

â‡’ an = 2 + [3 + 9 + 27 +â€¦ + (n- 1) term]

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

So, an = 2 + [3(3n-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