u-n-n-is-arithmetic-progression-u-n-u-0-nr-find-S-n-k-0-n-u-k-2-

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}]

Facebook Tweet Pin