Question

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

• A. $\frac{n(4n^2-1)c^2}{6}$
• B. $\frac{n(4n^2+1)c^2}{3}$
• C. $\frac{n(4n^2-1)c^2}{3}$
• D. $\frac{n(4n^2+1)c^2}{6}$

Answer 1

The correct option is C $\frac{n(4n^2-1)c^2}{3}$

sum of $n$ terms $=c{n}^2$

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

$0+2cn=(2a-d)+dn$

On company,

$2c=d$ and $2a=d$

$\Rightarrow d=2c$ and $a=c$

$\Rightarrow AP$ is $[c,c+2c,c+4c,.......,(2n-2)c+c]$

$\Rightarrow [c,3c,5c,...(2n-1\left)c\right]$

sum of squares $\Rightarrow [c^2+(3c{)}^2+(5c{)}^2+.......(2c-1{)}^2{c}^2]$

sum $\Rightarrow c^2[1^2+3^2+5^2+7^2+.....(2n-1{)}^2]$

Consider,

$[1^2+3^2+5^2+7^2|+.......(2n-1{)}^2]$

$=\sum (2n{)}^2-2^2-4^2......-(2n{)}^2$

$=\frac{\left(2n\right)(2n+1)(4n+1)}{6}-2^2[1^2+2^2+.....n^2]$

$=\frac{n(2n+1)(4n+1)}{3}-2\left[\frac{n(n+1)(2n+1)}{3}\right]$

$=\frac{n(2n+1)}{3}\left[\right(4n+1)-2(n+1\left)\right]$

$=\frac{n(2n+1)}{3}[2n-1]$

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

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