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 [2009]

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 [2009]

Smart Achievers · Accepted Answer

(3)

$Given: For an A.P., S_{n} = c n^{2}$

$Then T_{n} = S_{n} - S_{n - 1} = c n^{2} - c {(n - 1)}^{2}$

$= (2 n - 1) c$

$∴$ $Sum of squares of n terms of this A.P.$

$= \sum T_{n}^{2} = \sum {(2 n - 1)}^{2} c^{2}$

$= c^{2} [4 \sum n^{2} - 4 \sum n + n]$

$= c^{2} [\frac{4 n (n + 1) (2 n + 1)}{6} - \frac{4 n (n + 1)}{2} + n]$

$= c^{2} n [\frac{2 (2 n^{2} + 3 n + 1) - 6 (n + 1) + 3}{3}]$

$= c^{2} n [\frac{4 n^{2} - 1}{3}]$

$= \frac{n (4 n^{2} - 1) c^{2}}{3}$

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