Question
In the sum of first $$n$$ terms of an A.P. is $$c{n^2}$$, then the sum of squares of these $$n$$ terms is
A.
$$\frac{{n\left( {4{n^2} - 1} \right){c^2}}}{6}$$
B.
$$\frac{{n\left( {4{n^2} + 1} \right){c^2}}}{3}$$
C.
$$\frac{{n\left( {4{n^2} - 1} \right){c^2}}}{3}$$
D.
$$\frac{{n\left( {4{n^2} + 1} \right){c^2}}}{6}$$
Answer :
$$\frac{{n\left( {4{n^2} - 1} \right){c^2}}}{3}$$
Solution :
$$\eqalign{
& {\text{Given that for an A}}{\text{.P}}{\text{., }}{S_n} = c{n^2} \cr
& {\text{Then }}\,{T_n} = {S_n} - {S_{n - 1}} = c{n^2} - c{\left( {n - 1} \right)^2} \cr
& \,\,\,\,\,\, = \left( {2n - 1} \right)c \cr
& \therefore \,\,{\text{Sum of squares of }}n{\text{ terms of this A}}{\text{.P}}{\text{.}} \cr
& {\text{ = }}\sum {T_n^2 = \sum {{{\left( {2n - 1} \right)}^2}.{c^2}} } \cr
& = \,\,{c^2}\left[ {4\sum {{n^2} - 4} \sum {n + n} } \right] \cr
& = \,\,{c^2}\left[ {\frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)}}{2} + n} \right] \cr
& = \,\,{c^2}n\left[ {\frac{{2\left( {2{n^2} + 3n + 1} \right) - 6\left( {n + 1} \right) + 3}}{3}} \right] \cr
& = \,\,{c^2}n\left[ {\frac{{4{n^2} - 1}}{3}} \right] = \frac{{n\left( {4{n^2} - 1} \right){c^2}}}{3} \cr} $$