JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 2)

If the sum of first $$n$$ terms of an A.P. is $$c{n^2}$$, then the sum of squares of these $$n$$ terms is

$${{n\left( {4{n^2} - 1} \right){c^2}} \over 6}$$

$${{n\left( {4{n^2} + 1} \right){c^2}} \over 3}$$

$${{n\left( {4{n^2} - 1} \right){c^2}} \over 3}$$

$${{n\left( {4{n^2} + 1} \right){c^2}} \over 6}$$

We have $${t_n} = c\{ {n^2} - {(n - 1)^2}\} $$

$$ = c(2n - 1)$$

$$ \Rightarrow t_n^2 = {c^2}(4{n^2} - 4n + 1)$$

$$ \Rightarrow \sum\limits_{n = 1}^n {t_n^2 = {c^2}\left\{ {{{4n(n + 1)(2n + 1)} \over 6} - {{4n(n + 1)} \over 2} + n} \right\}} $$

$$ = {{{c^2}n} \over 6}\{ 4(n + 1)(2n + 1) - 12(n + 1) + 6\} $$

$$ = {{{c^2}n} \over 3}\{ 4{n^2} + 6n + 2 - 6n - 6 + 3\} = {{{c^2}} \over 3}n(4{n^2} - 1)$$

which is the sum of the square of $$n$$ terms.