Verified Solution • 2025–26 Session moderate

QJEEM23Macbb2: JEE Main 2023 Mathematics - Class 11 JEEM23Macbb2 Solution: Let and ∑ j = 1 n S j 2 = n A ( B n 2 + C n + D ) , where and…

Let SK=1+2+…+KKS_{K}=\frac{1+2+\ldots+K}{K} and ∑j=1nSj2=nA(Bn2+Cn+D), where A,B,C,D∈NA, B, C, D \in \mathbb{N} and AA has least value. Then

Multiple Choice Options

  • A A+B+C+DA+B+C+D is divisible by 5
  • B A+C+DA+C+D is not divisible by BB
  • C A+B=5(D−C)A+B=5(D-C)
  • D A+BA+B is divisible by D\mathrm{D}

Step-by-Step Verified Solution

amp;∵Sk=1+2+…+kkamp;=k(k+1)2k=k+12\begin{aligned}& \because S_k=\frac{1+2+\ldots+k}{k} \\\\& =\frac{k(k+1)}{2 k}=\frac{k+1}{2}\end{aligned}

amp;⇒Sk2=(k+12)2=k2+1+2k4amp;⇒∑j=1nSj2=14[∑j=1nk2+∑j=1n1+2∑j=1nk]\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}

amp;=14[n(n+1)(2n+1)6+n+2n(n+1)2]amp;=n4[(n+1)(2n+1)6+1+n+1]amp;=n24[2n2+3n+1+6+6n+6]amp;=n24[2n2+9n+13]\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

A=24, B=2,C=9,D=13\mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13

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

(B) A+C+D=46\mathrm{A}+\mathrm{C}+\mathrm{D}=46 , which is divisible by 2(B).

(C) A+B=26A+B=26

amp;5(D−C)=5(13−9)=20amp;∴26≠20\begin{aligned}& 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\& \therefore 26 \neq 20\end{aligned}

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

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