The sum of n terms of an AP is 4n(n-1),then what is the sum of their squares

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{S}}_{\mathrm{n}}=4\mathrm{n}\left(\mathrm{n}-1\right) \\ \mathrm{If} {\mathrm{T}}_{\mathrm{n}} \mathrm{denotes} {\mathrm{n}}_{\mathrm{th}} \mathrm{term}, \mathrm{then} \\ {\mathrm{T}}_1={\mathrm{S}}_1 \\ \Rightarrow {\mathrm{T}}_1=4\times 1\left(1-1\right) \\ \Rightarrow {\mathrm{T}}_1=0 \\ {\mathrm{S}}_2={\mathrm{T}}_1+{\mathrm{T}}_2 \\ \Rightarrow 4\times 2\left(2-1\right)=0+{\mathrm{T}}_2 \\ \Rightarrow {\mathrm{T}}_2=8 \\ \mathrm{Hence} \mathrm{A}.\mathrm{P}. \mathrm{is} 0,8,16,..... \\ {\mathrm{T}}_{\mathrm{n}}=0+\left(\mathrm{n}-1\right)8 \\ {\mathrm{T}}_{\mathrm{n}}=8\left(\mathrm{n}-1\right) \\ \mathrm{We} \mathrm{need} \\ \mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{n}} {\left[{\mathrm{T}}_{\mathrm{n}}\right]}^2 \\ =64\sum _{\mathrm{n}=1}^{\mathrm{n}} {\left[\left(\mathrm{n}-1\right)\right]}^2 \\ =64\sum _{\mathrm{n}=1}^{\mathrm{n}} \left[{\mathrm{n}}^2-2\mathrm{n}+1\right] \\ =64 \left[\sum _{\mathrm{n}=1}^{\mathrm{n}}{\mathrm{n}}^2-2\sum _{\mathrm{n}=1}^{\mathrm{n}}\mathrm{n}+\sum _{\mathrm{n}=1}^{\mathrm{n}}1\right] \\ =64\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-2\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\right] \\ =64\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-\mathrm{n}\left(\mathrm{n}+1\right)+\mathrm{n}\right] \end{gathered}$$

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