If-the-nth-term-of-a-sequence-is-given-by-n-2-2n-4-what-is-the-sum-of-n-terms-of-the-sequence-

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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