Sequences and Series 2 Question 8
9.
If the sum of first $n$ terms of an $AP$ 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) $\frac{n\left(4 n^{2}+1\right) c^{2}}{6}$
(2009)
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Answer:
Correct Answer: 9. (c)
Solution:
- Let $S _n=c n^{2}$
$ S _{n-1}=c(n-1)^{2}=c n^{2}+c-2 c n $
$T _n=2 c n-c \quad\left [\because T _n=S _n-S _{n-1}\right]$
$T _n^{2}=(2 c n-c)^{2}=4 c^{2} n^{2}+c^{2}-4 c^{2} n$
Sum= $\in T_n^2 = \frac{4c^2.n(n+1)(2n+1)}{6} + nc^2 -2c^2n(n+1)$
$=\frac{2c^2n(n+1) (2n+1) + 3nc^2 -6c^2n(n+1)}{3} $
$= \frac{nc^2(4n^2 +6n +2 +3 -6n -6) }{3}$
$=\frac{nc^2(4n^2-1)}{3}$
