# Sum of n Terms of Special Series

## Sum of n Terms of Special Series

Sum of n Terms of Special Series ∑n, ∑n² and ∑n³:

Generally, find the sum of n terms of any series, we use ∑ before the nth term of the series.

Sum of some special series is given below.

(i) Sum of n natural numbers = 1 + 2 + 3 + … + n

∑n = n(n + 1)/2

(ii) Sum of squares of n natural numbers = 1² + 2² + 3² + … + n²

$$\sum {{n}^{2}}\,=\,\frac{n(n+1)(2n+1)}{2}$$.

(iii) Sum of cubes of n natural numbers =1³ + 2³ + 3³ + … + n³

$$\sum {{n}^{3}}\,=\,\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$$.

Example: The sum of the n terms of the series 1² + 3² + 5² + … to n terms is

Solution: Given that,

1² + 3² + 5² + …

1, 3, 5 are in Ap

First term a = 1 and d = 3 – 1 = 2

Nth term tn = [1 + (n – 1) x 2] ²

= (2n – 1)²

= 4n² + 1 – 4n

$${{S}_{n}}\,=\,\sum\limits_{k=1}^{n}{(4{{k}^{2}}\,+\,1\,-\,4k)}$$ .

$$=\,4\sum\limits_{k=1}^{n}{{{k}^{2}}\,-\,4}\sum\limits_{k=1}^{n}{k}\,+\,n$$.

= 4(⅙) n(n + 1) (2n + 1) – 4. ½ n(n + 1) + n

= n/3 [ 2(n + 1) (2n + 1) – 6(n + 1) + 3]

= n/3(4n² – 1).