If first n terms of an A.P. is cn^2 , find sum of squares of its first n terms.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 109901 by ajfour last updated on 26/Aug/20

$I f f i r s t n t e r m s o f a n A . P . i s {c n}^{2},$ $f i n d s u m o f s q u a r e s o f i t s f i r s t$ $n t e r m s .$
	Answered by mr W last updated on 26/Aug/20

	$a_{1} + (a_{1} + d) + (a_{1} + 2 d) + . . . . + (a_{1} + (n - 1) d)$ $= {n a}_{1} + \frac{n (n - 1) d}{2} = n [a_{1} - \frac{d}{2} + \frac{n d}{2}] = {c n}^{2}$ $\Rightarrow a_{1} - \frac{d}{2} = 0 \Rightarrow d = 2 a_{1}$ $\Rightarrow \frac{d}{2} = c \Rightarrow d = 2 c$ $\Rightarrow a_{1} = c, d = 2 c$ $c^{2} + {(3 c)}^{2} + {(5 c)}^{2} + . . . {((2 n - 1) c)}^{2}$ $= [1^{2} + 3^{2} + 5^{2} + . . . {(2 n - 1)}^{2}] c^{2}$ $= \frac{n (4 n^{2} - 1) c^{2}}{3}$
	Commented by ajfour last updated on 26/Aug/20

	$M a r v e l l o u s S i r, t o o g o o d!$ $b u t h o w c o u l d y o u k n o w t h e$ $a n s w e r a f t e r t h e 2^{n d} l a s t s t e p,$ $S i r ?$
	Commented by mr W last updated on 26/Aug/20

	$\underset{k = 1}{\sum^{n}} {(2 k - 1)}^{2}$ $= \underset{k = 1}{\sum^{n}} (4 k^{2} - 4 k + 1)$ $= 4 \times \frac{n (n + 1) (2 n + 1)}{6} - 4 \times \frac{n (n + 1)}{2} + n$ $= \frac{4 (n - 1) n (n + 1)}{3} + n$ $= (\frac{4 n^{2} - 4 + 3}{3}) n$ $= \frac{n (4 n^{2} - 1)}{3}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum