If $t_{n} = (n + 4) (n + 5) (n + 6)$ .Find $S_{n}$

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Solution

$S_{n} = \sum (n + 4) (n + 5) (n + 6)$

$= \sum n^{3} + (4 + 5 + 6) n^{2} + (4 \times 5 + 5 \times 6 + 6 \times 4) n + 120$

$= \sum n^{3} + 15 n^{2} + (20 + 30 + 24) n + 120$

$= \sum n^{3} + 15 n^{2} + 74 n + 120$

$= \sum n^{3} + 15 \sum n^{2} + 74 \sum n + 120 \sum 1$

$= \frac{n^{2} {(n + 1)}^{2}}{4} + 15 \frac{n (n + 1) (2 n + 1)}{6} + 74 \frac{n (n + 1)}{2} + 120 n$

$= \frac{n^{2} (n^{2} + 2 n + 1)}{4} + 15 \frac{n (2 n^{2} + 3 n + 1)}{6} + 74 \frac{n (n + 1)}{2} + 120 n$

$= \frac{n^{4} + 2 n^{3} + n^{2}}{4} + 5 \frac{2 n^{3} + 3 n^{2} + n}{2} + 74 \frac{n^{2} + n}{2} + 120 n$

$= \frac{n^{4} + 2 n^{3} + n^{2}}{4} + \frac{10 n^{3} + 15 n^{2} + 5 n}{6} + 37 n^{2} + 37 n + 120 n$

$= \frac{6 n^{4} + 12 n^{3} + 6 n^{2} + 40 n^{3} + 60 n^{2} + 20 n + 888 n^{2} + 888 n + 4440 n}{24}$

$= \frac{3 n^{4} + 26 n^{3} + 477 n^{2} + 2674 n}{12}$

$= \frac{n}{12} (3 n^{3} + 26 n^{2} + 477 n + 2674)$

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