Question

If the sum of n terms of an A.P is cn(n - 1), where $c>0$, the sum of the square of these terms is -

• A. $c^2{n}^2(n+1{)}^2$
• B. $\frac{2}{3}{c}^{2}n(n-1)(2n-1)$
• C. $\frac{2c^2}{3}n(n-1)(2n-1)$
• D. none of these

Answer 1

The correct option is B

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

Solution : $S_n=cn(n-a);c>0$

$n=1$

$T_1=0$

$S_2=2c$

$T_2=2c$

$S_3=6c$

$T_3=4c$

$T_4=6c$

$s=0^2+(2c{)}^2+(4c{)}^2+(6c{)}^2+......+$

$S_{n-1}=4(c^2+4c^2+9c^2......t_{n-1})$

$S_{n-1}=4(c^2+4c^2+......t_{n-2}+t_{n-1})$

$t_{n-1}=4(c^2+3c^2+5c^2+........)$

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

$t_r=4c^2{r}^r$

Sum of squares
$\sum _{r=1}^{n-1}4c^2{r}^2=4c^2.\frac{(n-1)n\left(\right)2n-1)}{6}$
$=\frac{2}{3}{c}^{2}n(n-1)(2n-1)$