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Mathematics · Sequence and Series

JEE Main 2025 — 4 April, Evening Shift — Question 32

If the sum of the first 20 terms of the series

4⋅14+3⋅12+14+4⋅24+3⋅22+24+4⋅34+3⋅32+34+\frac{4 \cdot 1}{4+3 \cdot 1^{2}+1^{4}}+\frac{4 \cdot 2}{4+3 \cdot 2^{2}+2^{4}}+\frac{4 \cdot 3}{4+3 \cdot 3^{2}+3^{4}}+ 4⋅44+3⋅42+44+…\frac{4 \cdot 4}{4+3 \cdot 4^{2}+4^{4}}+\ldots is mn\frac{m}{n}, where mm and nn are

coprime, then m+nm+n is equal to :

  1. Option A:

    420

  2. Option B:

    423

  3. Option C:

    421

    Correct
  4. Option D:

    422

Answer: C

Step-by-step solution

Sn=∑r=1n4r4+3r2+r4S_{n}=\sum_{r=1}^{n} \frac{4 r}{4+3 r^{2}+r^{4}}

=2∑r=1n2r(r2+2)2−r2=2∑r=1n(r2+2+r)−(r2+2−r)(r2+2+r)(r2+2−r)=2∑r=1n(1r2+2−r−1r2+2+r)S20=2[(12−14)+(14−18)+…]=2(12−1202+2+20)=2(12−1422)=2(422−2422×2)=420422=210211=mnm+n=421\begin{aligned} & =2 \sum_{r=1}^{n} \frac{2 r}{\left(r^{2}+2\right)^{2}-r^{2}}=2 \sum_{r=1}^{n} \frac{\left(r^{2}+2+r\right)-\left(r^{2}+2-r\right)}{\left(r^{2}+2+r\right)\left(r^{2}+2-r\right)} \\& =2 \sum_{r=1}^{n}\left(\frac{1}{r^{2}+2-r}-\frac{1}{r^{2}+2+r}\right) \\& S_{20}=2\left[\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\ldots\right] \\& \quad=2\left(\frac{1}{2}-\frac{1}{20^{2}+2+20}\right) \\& \quad=2\left(\frac{1}{2}-\frac{1}{422}\right) \\& \quad=2\left(\frac{422-2}{422 \times 2}\right)=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \\& m+n=421 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Telescopic Summation

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