Question

Find the sum to $n$ terms of the series, whose $n$th term is given by $n^2+2^n$.

Answer 1

Given $a_n=n^2+2^n$

Now, sum of n terms is

$S_n={\sum }_{n=1}^n{a}_n$

$={\sum }_{n=1}^n{n}^2+2^n$

$={\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^2{2}^n.......\left(1\right)$

Now,

${\sum }_{n=1}^n{2}^n=2^1+2^2+2^3+.....+2^n$

$=2+4+8+....+2^n$

This is GP with first term $a=2$

And common ratio $r=\frac{4}{2}=2$

We know

$S_n=\frac{a(r^n-1)}{r-1}$

${\sum }_{n=1}^n{2}^n=\frac{2(2^n-1)}{2-1}$

${\sum }_{n=1}^n{2}^n=2(2^n-1)$

We also know

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

Now form (1)

$S_n={\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^2{2}^n$

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