Mathematics - Sequence & Series Question with Solution | TestHub

Let $S_n=c^2$ $$\begin{array}{rl}& S_{n-1}=c(n-1{)}^2=c{n}^2+c-2cn\\ & \therefore T_n=2cn-c\left(\because T_n=S_n-S_{n-1}\right)\\ & T_n^2=(2cn-c{)}^2=4c^2{n}^2+c^2-4c^{2}n\\ & \therefore \text{ Sum }=\Sigma T_n^2=\frac{4c^2\cdot n(n+1)(2n+1)}{6}\\ & =\frac{2c^{2}n(n+1)(2n+1)+3n{c}^2-6c^{2}n(n+1)}{3}\\ & =\frac{n{c}^2\left[4n^2+6n+2+3-6n-6\right]}{3}\\ & =\frac{n{c}^2\left(4n^2-1\right)}{3}\end{array}$$