SATHEE: Sequences and Series 2 Question 8

Sequences and Series 2 Question 8

9.

If the sum of first $n$ terms of an $A P$ is $c n^{2}$ , then the sum of squares of these $n$ terms is

(a) $\frac{n (4 n^{2} - 1) c^{2}}{6}$

(b) $\frac{n (4 n^{2} + 1) c^{2}}{3}$

(d) $\frac{n (4 n^{2} + 1) c^{2}}{6}$

(2009)

Show Answer

Answer:

Correct Answer: 9. (c)

Solution:

Let $S_{n} = c n^{2}$

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

$T_{n} = 2 c n - c [∵ T_{n} = S_{n} - S_{n - 1}]$

$T_{n}^{2} = (2 c n - c)^{2} = 4 c^{2} n^{2} + c^{2} - 4 c^{2} n$

Sum= $\in T_{n}^{2} = \frac{4 c^{2} . n (n + 1) (2 n + 1)}{6} + n c^{2} - 2 c^{2} n (n + 1)$

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

$= \frac{n c^{2} (4 n^{2} + 6 n + 2 + 3 - 6 n - 6)}{3}$

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

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Sequences and Series 2 Question 8

9.

Answer:

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