(u_n )_n is arithmetic progression/ u_n = u_0 +nr find S_n = Σ_(k=0) ^n u


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Question Number 29149 by abdo imad last updated on 04/Feb/18

${(u_{n})}_{n} i s a r i t h m e t i c p r o g r e s s i o n / u_{n} = u_{0} + n r$ $f i n d S_{n} = \sum_{k = 0}^{n} u_{k}^{2} .$
	Answered by mrW2 last updated on 04/Feb/18

	$u_{k} = u_{0} + k r$ $u_{k}^{2} = u_{0}^{2} + 2 u_{0} r k + r^{2} k^{2}$ $\underset{k = 0}{\sum^{n}} u_{k}^{2} = (n + 1) u_{0}^{2} + 2 u_{0} r \underset{k = 0}{\sum^{n}} k + r^{2} \underset{k = 0}{\sum^{n}} k^{2}$ $= (n + 1) u_{0}^{2} + 2 u_{0} r \times \frac{(n + 1) n}{2} + r^{2} \times \frac{n (n + 1) (2 n + 1)}{6}$ $= (n + 1) [u_{0}^{2} + {n u}_{0} r + \frac{n (2 n + 1) r^{2}}{6}]$
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