Question

Find the sum to n terms of the series whose nth term is given by

$n^2+2^n$

Answer 1

Let $a_n$ be the nth term and '$S_n$' be the sum of the n terms of the given series.

$\because a_n=n^2+2^n$ [Given]

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(k^2+2^k)$

$=(1^2+2^1)+(2^2+2^2)+(3^2+2^3)+\dots \dots +(n^2+2^n)$

$=(1^2+2^2+3^2+\dots \dots +n^2)+(2^1+2^2+2^3+\dots \dots +2^n)$

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

$=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)$.