Question

If $S_n=\lambda n(n-1)$ is an A.P. where $\lambda \ne 0$ then prove that the sum of the squares of the n terms of the A.P. is $\frac{2}{3}{\lambda }^{2}n(n-1)(2n-1)$.

Answer 1

$S_n=\lambda n(n-1)\therefore T_n=S_n-S_{n-1}$
$or{T}_n=\lambda n(n-1)-\lambda (n-1)(n-2)$
$or{T}_n=\lambda (n-1)\cdot 2$
$\sum T_n^2=4{\lambda }^2{\sum }_{n=1}^n(n-1{)}^2$
$=4{\lambda }^2[0^2+1^2+2^2+3^2+....+(n-1{)}^2]$
$=4{\lambda }^2\sum N^2=\frac{4{\lambda }^{2}N(N+1)(2N+1)}{6}WhereN=n-1$
$=\frac{4{\lambda }^2\cdot (n-1)n(2n-1)}{6}=\frac{2}{3}{\lambda }^{2}n(n-1)(2n-1)$