Let S_{k} = frac{1+2+ldots+k}{k} and sum_{j=1 | vedclass

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(A) Given $S_{k} = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.Then $S_{j}^2 = \frac{(j+1)^2}{4} = \frac{j^2 + 2j + 1}{4}$.Summing from $j=1$ to $n$:$\sum_{j=1

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$A + B$ is divisible by $D$

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(A) Given $S_{k} = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.Then $S_{j}^2 = \frac{(j+1)^2}{4} = \frac{j^2 + 2j + 1}{4}$.Summing from $j=1$ to $n$:$\sum_{j=1}^n S_j^2 = \frac{1}{4} \left[ \sum_{j=1}^n j^2 + 2 \sum_{j=1}^n j + \sum_{j=1}^n 1 ight]$$= \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} + n ight]$$= \frac{n}{4} \left[ \frac{(n+1)(2n+1)}{6} + (n+1) + 1 ight]$$= \frac{n}{24} \left[ (2n^2 + 3n + 1) + 6n + 6 + 6 ight]$$= \frac{n}{24} [2n^2 + 9n + 13]$.Comparing with $\frac{n}{A}(Bn^2 + Cn + D)$,we get $A=24, B=2, C=9, D=13$.Checking options:$A+B+C+D = 24+2+9+13 = 48$ (not divisible by $5$).$A+B = 26$,$D=13$,$26$ is divisible by $13$.$A+B = 26$,$5(D-C) = 5(13-9) = 20$. $26 
eq 20$.$A+C+D = 24+9+13 = 46$,$B=2$. $46$ is divisible by $2$.Thus,option $A$ is correct.

Let $S_{k} = \frac{1+2+\ldots+k}{k}$ and $\sum_{j=1}^n S_j^2 = \frac{n}{A}(Bn^2 + Cn + D)$,where $A, B, C, D \in \mathbb{N}$ and $A$ has the least value. Then:

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