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Find the sum of the first $n$ terms of the series whose nth term is $n^2+2^n$.
A$\frac{1}{6}n(n+1)(2n+1)+2^n-1$
B$\frac{1}{6}n(n+1)(2n+1)+2(2^n-1)$
C$\frac{1}{6}n(n+1)(2n+1)+2^{n+1}-1$
D$\frac{1}{6}n(n+1)(2n+1)+\frac{2^n-1}{2}$
Answer & Solution
Correct answer: B. $\frac{1}{6}n(n+1)(2n+1)+2(2^n-1)$
Here
$$S_n=\sum_{k=1}^n (k^2+2^k)=\sum_{k=1}^n k^2+\sum_{k=1}^n 2^k.$$The first sum is $\frac{n(n+1)(2n+1)}{6}$. The second is a geometric series with first term $2$, ratio $2$, and $n$ terms, so
$$\sum_{k=1}^n 2^k=2(2^n-1).$$Hence
$$S_n=\frac{1}{6}n(n+1)(2n+1)+2(2^n-1).$$Option A is tempting because it resembles the sum from $2^0$ to $2^{n-1}$, not from $2^1$ to $2^n$.