Find the sum to $n$ terms of the series whose $n^{t h}$ term is $n (n + 4)$ .

\frac{(n + 1) (n + 5)}{3}

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\frac{n (n + 1) (2 n + 13)}{6}

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\frac{n (n - 1) (n + 5)}{3}

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\frac{n (n + 1) (n + 5)}{6}

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Solution

The correct option is B $\frac{n (n + 1) (2 n + 13)}{6}$
Given that $a_{n} = n (n + 4) = n^{2} + 4 n$
Thus, the sum to n terms is given by
$S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} k^{2} = 4 \sum_{k = 1}^{n} k$

$= \frac{n (n + 1) (2 n + 1)}{6} + \frac{4 n (n + 1)}{2} = \frac{n (n + 1) (2 n + 13)}{6}$

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Find the sum to n terms of the series whose nth term is given by

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

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