JEE Advanced · Mathematics · 5. Sequences & Series
If the sum of first \(n\) terms of an \(\mathrm{AP}\) is \(\mathrm{cn}^2\), 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 & Solution
Correct Answer
(C) \(\frac{n\left(4 n^2-1\right) c^2}{3}\)
Step-by-step Solution
Detailed explanation
Let \(S_n=c^2\)
\(S_{n-1}=c(n-1)^2=c n^2+c-2 c n \)
\( \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \)
\( T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \)
\( \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \)
\( =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \)
\( =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \)
\( =\frac{n c^2\left(4 n^2-1\right)}{3}\)
\(S_{n-1}=c(n-1)^2=c n^2+c-2 c n \)
\( \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \)
\( T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \)
\( \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \)
\( =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \)
\( =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \)
\( =\frac{n c^2\left(4 n^2-1\right)}{3}\)
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