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).