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

n^th term of the series as:

a_n = n² + 2ⁿ

Then, the sum of n terms of the series can be expressed as

\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of first n terms of the given series is

\( S_n= \displaystyle\sum_{k=1}^{n}k^2+2^k\)

\( = \displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n}2^k\)

Let \( \displaystyle\sum_{k=1}^{n}2^k=2^1+2^2+2^3+...\)

Both the first term and the common ratio of \( 2^1+2^2+2^3+...\) which forms a G.P. is 2 .

Therefore,

\( \displaystyle\sum_{k=1}^{n}2^k=\dfrac{(2)[(2)^n-1]}{2-1}\)

= \(2 (2^n-1)\)

Then,

\( S_n= \displaystyle\sum_{k=1}^{n}k^2+2(2^n+1)\)

\( \dfrac{n(n+1)(2n+1)}{6}+2n(n+1)n\)

Therefore, the sum of n terms of the series whose ntg term is given by \( n^2+2^n\)

is \( \dfrac{n(n+1)(2n+1)}{6}+2n(n+1)n\)