JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 4 April 2025, Shift 2, Maths Q5

Question 5 of 75 in this shift, Maths question 5 of 25, Section A.

If the sum of the first 20 terms of the series

\frac{4\cdot1}{4+3\cdot1^2+1^4}+\frac{4\cdot2}{4+3\cdot2^2+2^4}+\frac{4\cdot3}{4+3\cdot3^2+3^4}+\frac{4\cdot4}{4+3\cdot4^2+4^4}+\ldots

\frac{m}{n}

, where m and n are coprime, then

m+n

is equal to :

(1)420
(2)421Official answer
(3)422
(4)423

Official answer

Option 2

NTA final key.

Chapter: Sequences and Series
Topic: AP and GP
Idea tested: Method of Differences for Telescoping Series

Same idea in other shifts

Asked 8× in all

5 Apr 2026, Shift 1 · Q5 $\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)$ is equal to:EasySingle correct
3 Apr 2025, Shift 1 · Q7The sum $1 + 3 + 11 + 25 + 45 + 71 + \ldots$ upto 20 terms, is equal toMediumSingle correct
9 Apr 2024, Shift 2 · Q24If…HardNumerical value
31 Jan 2024, Shift 1 · Q5The sum of the series $\frac{1}{1-3\cdot1^2+1^4}+\frac{2}{1-3\cdot2^2+2^4}+\frac{3}{1-3\cdot3^2+3^4}+\ldots$ up to 10-terms isMediumSingle correct
6 Apr 2023, Shift 1 · Q7The sum of the first 20 terms of the series $5+11+19+29+41+\ldots$ isMediumSingle correct
8 Apr 2023, Shift 2 · Q8Let $a_n$ be the $n^{th}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \ldots$ and $S_n = \sum_{k=1}^{n} a_k$ . Then $S_{30} - a_{40}$ …MediumSingle correct
18 Mar 2021, Shift 1 · Q68 $\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\ldots+\frac{1}{(201)^2-1}$ is equal to :MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.