∑ _ #160; K=1^ n (K+1)(3k 1) #160; ∑_K=1^n(3k^2+2k 1) ∑_K=1^n(3K^3+2K^3 k) ∑_K=1^n3k^3 + ∑_K=1^n2k^3 ∑_K=1^nk =3.n^2(n+1)^24 ...

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Integration Using Substitution

∑nk=1kk+13k-1...

Question

$n \sum k = 1 k (k + 1) (3 k - 1)$ .

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Solution

$n \sum K = 1 (K + 1) (3 k - 1) n \sum K = 1 (3 k^{2} + 2 k - 1)$
$n \sum K = 1 (3 K^{3} + 2 K^{3} - k)$
$n \sum K = 1 3 k^{3}$ + $n \sum K = 1 2 k^{3}$ - $n \sum K = 1 k$
$
$= 3. \frac{n^{2} (n + 1)^{2}}{4} + 2 \times \frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1}{2}$
$= \frac{9 n^{2} (n + 1)^{2} + 4 n (n + 1) (2 n + 1) - 6 n (n + 1)}{12}$
$= \frac{n (n + 1) (9 n (n + 1) + 4 (2 n + 1) - 6}{12}$
$= \frac{n (n + 1) (9 n^{2} + 17 n - 2)}{12}$
$= \frac{n (n + 1) (9 n^{2} + 18 n - n - 2}{)} 12$
$= \frac{n (n + 1) (9 n - 1) (n + 2)}{12}$

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