JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 14)
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
$$A+B+C+D$$ is divisible by 5
$$A+C+D$$ is not divisible by $$B$$
$$A+B=5(D-C)$$
$$A+B$$ is divisible by $$\mathrm{D}$$
Explanation
$$
\begin{aligned}
& \because S_k=\frac{1+2+\ldots+k}{k} \\\\
& =\frac{k(k+1)}{2 k}=\frac{k+1}{2}
\end{aligned}
$$
$$ \begin{aligned} & \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\ & \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right] \end{aligned} $$
$$ \begin{aligned} & =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\\\ & =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\\\ & =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\\\ & =\frac{n}{24}\left[2 n^2+9 n+13\right] \end{aligned} $$
On comparing, we get
$$ \mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13 $$
(A) $A+B+C+D=48$, which is not divisible by 5.
(B) $\mathrm{A}+\mathrm{C}+\mathrm{D}=46$, which is divisible by 2(B).
(C) $A+B=26$
$$ \begin{aligned} & 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\ & \therefore 26 \neq 20 \end{aligned} $$
(D) $A+B=24+2=26$, divisible by 13.
$$ \begin{aligned} & \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\ & \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right] \end{aligned} $$
$$ \begin{aligned} & =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\\\ & =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\\\ & =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\\\ & =\frac{n}{24}\left[2 n^2+9 n+13\right] \end{aligned} $$
On comparing, we get
$$ \mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13 $$
(A) $A+B+C+D=48$, which is not divisible by 5.
(B) $\mathrm{A}+\mathrm{C}+\mathrm{D}=46$, which is divisible by 2(B).
(C) $A+B=26$
$$ \begin{aligned} & 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\ & \therefore 26 \neq 20 \end{aligned} $$
(D) $A+B=24+2=26$, divisible by 13.
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