If-first-n-terms-of-an-A-P-is-cn-2-find-sum-of-squares-of-its-first-n-terms-

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) + \dots . + (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} + \dots {((2 n - 1) c)}^{2}

= [1^{2} + 3^{2} + 5^{2} + \dots {(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}

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