First slide

Sequence and Series

Question

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

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} ∵ S_{n} = c n^{2} \\ ∴ t_{n} = S_{n} - S_{n - 1} = c (2 n - 1) \\ Σ t_{n}^{2} = c^{2} Σ (2 n - 1)^{2} \\ = c^{2} Σ (4 n^{2} - 4 n + 1) = c^{2} \{4 Σ n^{2} - 4 Σ n + Σ 1\} \\ = c^{2} \{\frac{4 n (n + 1) (2 n + 1)}{6} - \frac{4 n (n + 1)}{2} + n\} \end{array}$
$\begin{array}{l} = c^{2} n \{\frac{2}{3} (2 n^{2} + 3 n + 1) - 2 n - 2 + 1\} \\ = \frac{c^{2} n}{3} (4 n^{2} - 1) = \frac{n (4 n^{2} - 1) c^{2}}{3} \end{array}$

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