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Question Number 113015 by mohssinee last updated on 10/Sep/20

	Answered by Olaf last updated on 10/Sep/20

	$\underset{k = 0}{\sum^{n}} (2 k^{2} + 3 k + 5) =$ $2 \underset{k = 0}{\sum^{n}} k^{2} + 3 \underset{k = 0}{\sum^{n}} k + \underset{k = 0}{\sum^{n}} 5 =$ $2 \frac{n (n + 1) (2 n + 1)}{6} + 3 \frac{n (n + 1)}{2} + 5 (n + 1)$ $= (n + 1) [\frac{n (2 n + 1)}{3} + \frac{3}{2} n + 5] =$ $(n + 1) [\frac{2}{3} n^{2} + \frac{1}{3} n + \frac{3}{2} n + 5] =$ $(n + 1) [\frac{2}{3} n^{2} + \frac{11}{6} n + 5] =$ $\frac{1}{6} (n + 1) (4 n^{2} + 11 n + 30)$
	Answered by mathmax by abdo last updated on 10/Sep/20
	$Σ_(k=0) ^n (2k^2 +3k +5) =2Σ_(k=0) ^n k^2 +3Σ_(k=0) ^n k +5Σ_(k=0) ^n (1) =2×((n(n+1)(2n+1))/6) +3.((n(n+1))/2) +5(n+1) =(n+1){((n(2n+1))/3) +((3n)/2) +5} =(n+1)( ((2n(2n+1)+9n +30)/6)) =((n+1)/6)(4n^2 +2n+9n +30) =(((n+1)(4n^2 +11n +30))/6)$
	$\sum_{k = 0}^{n} ({2 k}^{2} + 3 k + 5) = 2 \sum_{k = 0}^{n} k^{2} + 3 \sum_{k = 0}^{n} k + 5 \sum_{k = 0}^{n} (1)$ $= 2 \times \frac{n (n + 1) (2 n + 1)}{6} + 3 . \frac{n (n + 1)}{2} + 5 (n + 1)$ $= (n + 1) {\frac{n (2 n + 1)}{3} + \frac{3 n}{2} + 5}$ $= (n + 1) (\frac{2 n (2 n + 1) + 9 n + 30}{6}) = \frac{n + 1}{6} ({4 n}^{2} + 2 n + 9 n + 30)$ $= \frac{(n + 1) ({4 n}^{2} + 11 n + 30)}{6}$
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