Question

If the sum of first $n$ terms of an A.P. is $\frac{n^2}{\sqrt{3}},$ then the sum of squares of first $18$ terms is

Answer 1

$S_n=\frac{n^2}{\sqrt{3}}$
$S_{n-1}=\frac{(n-1{)}^2}{\sqrt{3}}$
$t_n=S_n-S_{n-1}=\frac{2n-1}{\sqrt{3}}$

Now,
$\sum _{n=1}^n{t}_n^2={\left(\frac{1}{\sqrt{3}}\right)}^2[1^2+3^2+...+(2n-1{)}^2]$
$=\frac{1}{3}\left[\right\{1^2+2^2+...+\left(2n{)}^2\right\}-\{2^2+4^2+...+(2n{)}^2\left\}\right]$
$=\frac{1}{3}[\frac{\left(2n\right)(2n+1)(4n+1)}{6}-2^2(1^2+2^2+...+n^2\left)\right]$
$=\frac{1}{3}[\frac{\left(2n\right)(2n+1)(4n+1)}{6}-\frac{\left(4n\right)(n+1)(2n+1)}{6}]$
$=\frac{1}{3}\times \frac{2n+1}{3}[n(4n+1)-2n(n+1)]$
$=\frac{n(4n^2-1)}{9}$

For $n=18$
$\sum _{n=1}^{18}{t}_n^2=\frac{18(4\times {18}^2-1)}{9}=2590$