$2 \times 3 + 3 \times 4 + 4 \times 5 + . . . . n t e r m s =$

\frac{n (n^{2} + 6 n + 11)}{6}

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\frac{n (n^{2} + 6 n + 14)}{9}

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\frac{n (n^{2} + 6 n + 11)}{3}

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\frac{n (n^{2} + 6 n + 17)}{12}

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Solution

The correct option is C $\frac{n (n^{2} + 6 n + 11)}{3}$
$2.3 + 3.4 + 4.5 + . . . . n t e r m s = n \sum r = 1 (r + 1) (r + 2) = n \sum r = 1 r^{2} + 3 r + 2 = n \sum r = 1 r^{2} + 3 n \sum r = 1 r + 2 n = \frac{n (n + 1) (2 n + 1)}{6} + 3 \frac{n (n + 1)}{2} + 2 n = \frac{n (2 n^{2} + 12 n + 22)}{6} = \frac{n}{3} (n^{2} + 6 n + 11)$

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