The sum to $n$ terms of an AP is an ( $n-1$ ). Then the sum of squares of the $n$ terms is

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Professor PC Thomas Classes · Answer

$ \begin{aligned} & S_n=a_n(n-1) \ & t_n=S_n-S_{n-1}=a[n(n-1)-(n-1)(n-2)] \ & =a(n-1)[2]=2 a(n-1) \ & S_n^{\prime}=\sum 4 a^2(n-1)^2 \quad d=2 \ & =4 a^2\left\{\frac{(n-1)(n)(2 n-1)}{6}ight\} \ & =\frac{2 a^2}{3}(n(n-1)(2 n-1)) \end{aligned} $