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Mathematics · Sequence and Series

JEE Main 2025 — 4 April, Morning Shift — Question 20

1+3+52+7+92+,…1+3+5^{2}+7+9^{2}+, \ldots upto 40 terms is equal to

  1. Option A:

    33980

  2. Option B:

    41880

    Correct
  3. Option C:

    40870

  4. Option D:

    43890

Answer: B

Step-by-step solution

1+3+52+7+92+…1+3+5^{2}+7+9^{2}+\ldots upto 40 terms

(12+52+92+…)+(3+7+11+…)=(∑k=120(4k−3)2)+202[6+(20−1)4]=16∑k=120k2−24∑k=120k+9×20+10[82]=16(20×21×416)−24(20×212)+1000=45920−5040+1000=41880\begin{aligned} & \left(1^{2}+5^{2}+9^{2}+\ldots\right)+(3+7+11+\ldots) \\& =\left(\sum_{k=1}^{20}(4 k-3)^{2}\right)+\frac{20}{2}[6+(20-1) 4] \\& =16 \sum_{k=1}^{20} k^{2}-24 \sum_{k=1}^{20} k+9 \times 20+10[82] \\& =16\left(\frac{20 \times 21 \times 41}{6}\right)-24\left(\frac{20 \times 21}{2}\right)+1000 \\& =45920-5040+1000 \\& =41880 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Introduction to Sequence and Series

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