Let [S_K=(1+2++K)/(K)] and [_j=1^n S_j^2=(n)/(A)(B n^2+C n+D)], where [A, B, C, D N] and A…

Question

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

Accepted Answer

$$
\begin{aligned}
& \because S_k=\frac{1+2+\ldots+k}{k} \\
& =\frac{k(k+1)}{2 k}=\frac{k+1}{2}
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\
& \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right]
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\
& =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\
& =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\
& =\frac{n}{24}\left[2 n^2+9 n+13\right]
\end{aligned}
$$

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.

Solutions · 1