Find the sum to n terms of the series whose n | vedclass

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(A) Given the $n^{th}$ term is $a_n = n^2 + 2^n$. The sum of $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 2^k)$. This can be split into

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

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(A) Given the $n^{th}$ term is $a_n = n^2 + 2^n$. The sum of $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 2^k)$. This can be split into two separate summations: $S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n 2^k$. The sum of the first $n$ squares is given by the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. The second part is a geometric progression: $\sum_{k=1}^n 2^k = 2^1 + 2^2 + \dots + 2^n$. Using the sum formula for a $G.P.$ with first term $a=2$ and common ratio $r=2$,we get $\frac{a(r^n - 1)}{r-1} = \frac{2(2^n - 1)}{2-1} = 2(2^n - 1) = 2^{n+1} - 2$. Combining these,$S_n = \frac{n(n+1)(2n+1)}{6} + 2(2^n - 1)$.

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $a_n = n^2 + 2^n$.

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