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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Sequences and Series

1+3+5^2+7+9^2+ upto 40 terms is equal to

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

Options

  1. A43890
  2. B41880
  3. C33980
  4. D40870

Correct answer

B. 41880

Step-by-step solution

(12+52+92+…… upto 20 terms )+(3+7+11+\left(1^2+5^2+9^2+\ldots \ldots \text{ upto } 20 \text{ terms }\right)+(3+7+11+...upto 20 terms)=∑r=120(4r−3)2+∑r=120(4r−1)=∑r=120(4r−3)2+(4r−1)=4∑r=120(4r2−5r+2)=16∑r=120r2−20∑r=120r+8∑r=1201=41880\begin{aligned}& =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-1) \\ & =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+(4 \mathrm{r}-1) \\ & =4 \sum_{\mathrm{r}=1}^{20}\left(4 \mathrm{r}^2-5 \mathrm{r}+2\right) \\ & =16 \sum_{\mathrm{r}=1}^{20} \mathrm{r}^2-20 \sum_{\mathrm{r}=1}^{20} \mathrm{r}+8 \sum_{\mathrm{r}=1}^{20} 1=41880\end{aligned}
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