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

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Question

(A) $a_n = n(n+1)(n+4) = n(n^2 + 5n + 4) = n^3 + 5n^2 + 4n$$	herefore S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n k^3 + 5\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k

Vedclass · Accepted Answer

$\frac{n(n+1)(3n^2+23n+34)}{12}$

Vedclass · Answer

(A) $a_n = n(n+1)(n+4) = n(n^2 + 5n + 4) = n^3 + 5n^2 + 4n$$	herefore S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n k^3 + 5\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k$$= \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$$= \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 ight]$$= \frac{n(n+1)}{2} \left[ \frac{3n^2 + 3n + 20n + 10 + 24}{6} ight]$$= \frac{n(n+1)}{2} \left[ \frac{3n^2 + 23n + 34}{6} ight]$$= \frac{n(n+1)(3n^2 + 23n + 34)}{12}$

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $a_n = n(n+1)(n+4)$.

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