JEE Main 2024 — Sequences And Series Question with Solution
From: JEE Main 2024 (Online) 4th April Evening Shift
Question
The value of is
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Show full solutionCorrect option: A
Step-by-step explanation
\begin{aligned} & \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\ & \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)} \end{aligned}
\begin{aligned} & \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}{\sum_\limits{n=1}^{100} n^3+n^2} \\ & =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\ & =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301} \end{aligned}
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This is a previous-year question from JEE Main 2024, covering the Sequences And Series chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.