Question

If the sum of first n terms of an A.P. is ${cn}^2$, 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 $\frac{n({4n}^2-1)c^2}{3}$

Let $S_n$ be the sum of the first $n$ terms.

Then,$S_n=c{n}^2$

If we denote the $n$th term as $T_n$,then

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

$S={\sum }_{r=1}^n{T_r}^2$

$S={\sum }_{r=1}^n{\left[c\right(2r-1\left)\right]}^2$

$S=c^2{\sum }_{r=1}^n{(2r-1)}^2$

$S=c^2{\sum }_{r=1}^n{r}^2-c^2{\sum }_{r=1}^n{\left(2r\right)}^2$

$S=c^2{\sum }_{r=1}^{2n}{r}^2-{4c}^2{\sum }_{r=1}^n{r}^2$

Since,${\sum }_{r=1}^n{r}^2=n(n+1)(2n+1)6,$

therefore $S=c^2(\frac{2n(2n+1)(2\times 2n+1)}{6}-4\frac{n(n+1)(2n+1)}{6})$

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

$S=c^2\ast \frac{2n(2n+1)}{6}\left[\right(4n+1)-2(n+1\left)\right]=\frac{2n(2n+1)(2n-1)c^2}{6}$ = $\frac{n({4n}^2-1)c^2}{3}$