Question

Find the sum of the series whose nth term is (3n^2-3n+2)

Answer 1

This is the answer given below:

Answer 2

Answer:

$S_n=n(n^2+1))$

Step-by-step explanation:

Given : nth term is $(3n^2-3n+2)$

To Find :Find the sum of the series

Solution:

Nth term is $t_n=(3n^2-3n+2)$

Sum of the series  =  $S_n=\sum^n _{n=1} t_n$

So,  $S_n=\sum^n _{n=1} (3k^2-3k+2)$

$S_n=3\sum^n _{n=1}k^2-3\sum^n _{n=1}k+2\sum^n _{n=1}<strong> </strong>1$

$S_n=3\frac{n(n+1)(2n+1)}{6}-3\frac{n(n+1)}{2}+2n$

$S_n=\frac{n(n+1)(2n+1)}{2}-3\frac{n(n+1)}{2}+2n$

$S_n=\frac{n(n+1)}{2}(2n+1-3)+2n$

$S_n=\frac{n(n+1)}{2}(2n-2)+2n$

$S_n=n(n+1)(n-1)+2n$

$S_n=n(n^2-1+2)$

$S_n=n(n^2+1))$

Hence the sum of the series whose nth term is $(3n^2-3n+2)$ is $S_n=n(n^2+1))$