Let $S_{K}=\frac{1+2+\ldots+K}{K}$ and $\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$, where $A, B, C, D \in \mathbb{N}$ and $A$ has least value. Then

$$ \begin{aligned} & \because S_k=\frac{1+2+\ldots+k}{k} \\\\ & =\frac{k(k+1)}{2 k}=\frac{k+1}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\ & \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right] \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\\\ & =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\\\ & =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\\\ & =\frac{n}{24}\left[2 n^2+9 n+13\right] \end{aligned} $$ <br/><br/>On comparing, we get <br/><br/>$\mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13$ <br/><br/>(A) $A+B+C+D=48$, which is not divisible by 5. <br/><br/>(B) $\mathrm{A}+\mathrm{C}+\mathrm{D}=46$, which is divisible by 2(B). <br/><br/>(C) $A+B=26$ <br/><br/>$$ \begin{aligned} & 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\ & \therefore 26 \neq 20 \end{aligned} $$ <br/><br/>(D) $A+B=24+2=26$, divisible by 13.