Question

The sum of $n$ terms of an $A.P$. is a $n(n-1)$. The sum of the squares of these terms is

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

Answer 1

The correct option is C $\frac{2a^2}{3}n\left(n-1\right)\left(2n-1\right)$

Given sum of n terms of an A.P. $S_1=$ $an(n-1)$. ……. (1)

Then, sum of (n-1) terms of an A.P.

$S_2=a(n-1)(n-1-1)$

$S_2=a(n-1)(n-2)$ ……. (2)

Now, We know that,

$n^{th}$ term of an A.P.$=S_1-S_2$

$=an(n-1)-a(n-1)(n-2)$

$=a(n-1)[n-n+2]$

$=2a(n-1)$

Putting n=1,2,3,4,5………….

And became an A.P.$0,2a,4a,6a,8a.............$

Squaring this A.P. and We get,

$0,4a^2,16a^2,36a^2,64a^2..............$

According to given question.

Sum of sequence :-

$0+4a^2+16a^2+36a^2+64a^2+..............$

$S_n=4a^2[0+1+4+9+16+\mathrm{25.............}]$

$S_n=4a^2[1^2+2^2+3^2+4^2+5^2+.............]$

$S_n=4a^2[1^2+2^2+3^2+4^2+5^2+.............{(n-1)}^2]$ …….. (3)

We know that the sum of $n$ natural numbers $=\frac{n(n+1)(2n+1)}{6}$

Then, the sum of $(n-1)$ natural numbers

$=\frac{(n-1)(n+1-1)(2(n-1)+1)}{6}$

$=\frac{n(n-1)(2n-1)}{6}$

Now, Sum of sequence (3)

The sum of squares

$S_{(n-1)}=\frac{4a^{2}n(n-1)(2n-1)}{6}$

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

Hence it is complete solution.

Option $\left(C\right)$ is correct.