Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Question:

Find the sum to $n$ terms of the series whose $n^{\text {th }}$ terms is given by $n^{2}+2^{n}$

Solution:

$a_{n}=n^{2}+2^{n}$

$\therefore S_{n}=\sum_{k=1}^{n} k^{2}+2^{k}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 2^{k}$ (1)

Consider $\sum_{k=1}^{n} 2^{k}=2^{1}+2^{2}+2^{3}+\ldots$

The above series $2,2^{2}, 2^{3}, \ldots$ is a G.P. with both the first term and common ratio equal to 2 .

$\therefore \sum_{k=1}^{n} 2^{k}=\frac{(2)\left[(2)^{n}-1\right]}{2-1}=2\left(2^{n}-1\right)$ (2)

Therefore, from (1) and (2), we obtain 

$S_{n}=\sum_{k=1}^{n} k^{2}+2\left(2^{n}-1\right)=\frac{n(n+1)(2 n+1)}{6}+2\left(2^{n}-1\right)$

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