Let S _ K = (1 + 2 + . . . + K)/(K) and ∑ _ j = 1 ^ n S ^ 2 j =...

Given that ${S}_{K} = \dfrac{1 + 2 + . . . + K}{K}$

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

$⇒ {S}_{K} = \dfrac{K\left(K + 1\right)}{2K} = \dfrac{K + 1}{2}$

Now,

$\mathop{∑}_{j = 1}^{n}{\left({S}_{j}\right)}^{2} = \mathop{∑}_{j = 1}^{n}\dfrac{1}{4}\left({2}^{2} + {3}^{2} + {4}^{2} . . . {\left(n + 1\right)}^{2}\right)$

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

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

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

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

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

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

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

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

$⇒ \dfrac{A + B}{D} = \dfrac{26}{13} = 2$

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

Hence this is the correct option.