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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2009 · Paper 2 · Mathematics · Sequences and Series

If the sum of first n terms of an AP is cn^2 , then the sum of...

If the sum of first nn terms of an AP\mathrm{AP} is cn2\mathrm{cn}^2, then the sum of squares of these nn terms is

Options

  1. An(4n2−1)c26\frac{n\left(4 n^2-1\right) c^2}{6}
  2. Bn(4n2+1)c23\frac{n\left(4 n^2+1\right) c^2}{3}
  3. Cn(4n2−1)c23\frac{n\left(4 n^2-1\right) c^2}{3}
  4. Dn(4n2+1)c26\frac{n\left(4 n^2+1\right) c^2}{6}

Correct answer

C. n(4n2−1)c23\frac{n\left(4 n^2-1\right) c^2}{3}

Step-by-step solution

Let Sn=c2S_n=c^2Sn−1=c(n−1)2=cn2+c−2cn∴Tn=2cn−c(\becauseTn=Sn−Sn−1)Tn2=(2cn−c)2=4c2n2+c2−4c2n∴ Sum =\SigmaTn2=4c2⋅n(n+1)(2n+1)6=2c2n(n+1)(2n+1)+3nc2−6c2n(n+1)3=nc2[4n2+6n+2+3−6n−6]3=nc2(4n2−1)3\begin{aligned}& S_{n-1}=c(n-1)^2=c n²+c-2 c n \\ & \therefore T_n=2 c n-c \quad\left(\becauseT_n=S_n-S_{n-1}\right) \\ & T_n^2=(2 c n-c)^2=4 c² n²+c²-4 c² n \\ & \therefore \text{ Sum }=\SigmaT_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} \end{aligned}$$
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