JEE Main 2025 Apr 4 Shift 1, Mathematics Q11: Summation of Series

JEE Main2025Apr 4, Shift 1Mathematics

$1 + 3 + 5^{2} + 7 + 9^{2} + \dots$ upto 40 terms is equal to

A
43890
B
41880
C
33980
D
40870

Solution

The terms alternate: odd-indexed positions ( $1^{s t}, 3^{r d}, 5^{t h}, \dots$ ) carry squared odd numbers $1^{2}, 5^{2}, 9^{2}, \dots$ while even-indexed positions carry plain odd numbers $3, 7, 11, \dots$ . With 40 terms total, there are 20 of each type.

Squared part: $r = 1 \sum 20 (4 r - 3)^{2}$ . Linear part: $r = 1 \sum 20 (4 r - 1)$ .

$r = 1 \sum 20 (4 r - 3)^{2} = r = 1 \sum 20 (16 r^{2} - 24 r + 9) = 16 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 24 \cdot \frac{20 \cdot 21}{2} + 9 \cdot 20 = 45920 - 5040 + 180 = 41060$ .

$r = 1 \sum 20 (4 r - 1) = 4 \cdot \frac{20 \cdot 21}{2} - 20 = 840 - 20 = 820$ .

Total $= 41060 + 820 = 41880$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q11: Summation of Series

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