Let S_K=(1+2+ +K)/(K) and _j=1^nS_2^j=(n)/(A)(Bn^2+Cn+D), where A,B,C,D N and A has least …

Question

Let ${S}_{K}=\frac{1+2+\ldots +K}{K}$ and $\sum _{j=1}^{n}{S}_{2}^{j}=\frac{n}{A}(B{n}^{2}+Cn+D)$, where $A,B,C,D\in N$ and $A$ has least value. Then

Accepted Answer

$\begin{gathered} \because {S}_{k}=\frac{1+2+\ldots +k}{k} \ =\frac{k(k+1)}{2k}=\frac{k+1}{2} \end{gathered}$

$\begin{gathered} \Rightarrow {S}_{2}^{k}={(\frac{k+1}{2})}^{2}=\frac{{k}^{2}+1+2k}{4} \ \Rightarrow \sum _{j=1}^{n}{S}_{2}^{j}=\frac{1}{4}[\sum _{j=1}^{n}{k}^{2}+\sum _{j=1}^{n}1+2\sum _{j=1}^{n}k] \end{gathered}$

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

On comparing, we get

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

(A) $A+B+C+D=48$, which is not divisible by 5.

(B) $A+C+D=46$, which is divisible by 2(B).

(C) $A+B=26$

$\begin{gathered} 5(D-C)=5(13-9)=20 \ 	herefore 26
e 20 \end{gathered}$

(D) $A+B=24+2=26$, divisible by 13.

Solutions · 1