First slide

Series of natural numbers

Question

If the sum of n terms of an A.P, is cn (n-1) , where $c \neq 0$ , then sum of the squares of these terms is

Moderate

A

$c^{2} n {(n + 1)}^{2}$
B

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

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

$c^{2}$

Solution

$if t_{r} be the r^{t h} term of the A.P. then$
$\begin{array}{l} t_{r} = S_{r} - S_{r - 1} \\ = c r (r - 1) - c (r - 1) (r - 2) \\ = c (r - 1) (r - r + 2) = 2 c (r - 1) \\ We have, t_{1}^{2} + t_{2}^{2} + \dots . + t_{n}^{2} = 4 c^{2} (0^{2} + 1^{2} + 2^{2} + \dots . + (n - 1)^{2}) \\ = 4 c^{2} \frac{(n - 1) n (2 n - 1)}{6} = \frac{2}{3} c^{2} n (n - 1) (2 n - 1) \end{array}$

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