QJEEAdv09Ma813d: JEE Advanced 2009 Mathematics - Class 11 JEEAdv09Ma813d Solution: If the sum of first terms of an A.P. is , then the sum of…

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.