JEE Main 2023 — Sequences and Series Question with Solution

Given that $S_K=\frac{1+2+...+K}{K}$

We know that $1+2+3....+n=\frac{n\left(n+1\right)}{2}$

$\Rightarrow S_K=\frac{K\left(K+1\right)}{2K}=\frac{K+1}{2}$

Now,

$\sum _{j=1}^n{\left(S_j\right)}^2=\sum _{j=1}^n\frac{1}{4}\left(2^2+3^2+4^2...{\left(n+1\right)}^2\right)$

$=\sum _{j=1}^n\frac{1}{4}\left(1^2+2^2+3^2+4^2...{\left(n+1\right)}^2-1^2\right)$

$\Rightarrow \frac{1}{4}\left(\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}-1\right)$

$=\frac{1}{4}\left(\frac{\left(n^2+3n+2\right)\left(2n+3\right)-6}{6}\right)$

$=\frac{1}{4}\left(\frac{2n^3+6n^2+4n+3n^2+9n+6-6}{6}\right)$

$=\frac{1}{4}\left(\frac{2n^3+9n^2+13n}{6}\right)$

$=\frac{n}{24}\left(2n^2+9n+13\right)$

On comparing this with $\sum _{j=1}^n{S^2}_j=\frac{n}{A}\left(B{n}^2+Cn+D\right)$ we get,

$A=24, B=2, C=9, D=13$

$\Rightarrow \frac{A+B}{D}=\frac{26}{13}=2$

That means, $A+B$ is divisible by $D$.

Hence this is the correct option.