If the nth term of a sequence is given by ((n^2 −2n)/4) ,what is the sum of n terms of the sequence?


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Question Number 206993 by necx122 last updated on 03/May/24

$I f t h e n t h t e r m o f a s e q u e n c e i s$ $g i v e n b y \frac{n^{2} - 2 n}{4}, w h a t i s t h e s u m o f n$ $t e r m s o f t h e s e q u e n c e ?$
	Answered by Rasheed.Sindhi last updated on 03/May/24

	$\underset{k = 1}{\sum^{n}} \frac{k^{2} - 2 k}{4}$ $= \underset{k = 1}{\sum^{n}} (\frac{k^{2}}{4} - \frac{k}{2})$ $= \frac{1}{4} \underset{k = 1}{\sum^{n}} k^{2} - \frac{1}{2} \underset{k = 1}{\overset{n}{Σ}} k$ $= \frac{1}{4} (\frac{n (n + 1) (2 n + 1)}{6}) - \frac{1}{2} (\frac{n (n + 1)}{2})$ $= \frac{n (n + 1) (2 n + 1)}{24} - \frac{n (n + 1)}{4}$ $= n (n + 1) (\frac{2 n + 1}{24} - \frac{1}{4})$ $= n (n + 1) (\frac{2 n + 1 - 6}{24})$ $= \frac{1}{24} n (n + 1) (2 n - 5)$
	Commented by necx122 last updated on 04/May/24

	$T h a n k y o u s o m u c h s i r . A l s o, i s t h e r e$ $a w a y t o p r o v e w h a t s u m o f a s q u a r e$ $f u n c t i o n ?$
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