JEE Main 2026 Apr 5 Shift 2, Mathematics Q5: Summation of Series

JEE Main2026Apr 5, Shift 2Mathematics

If the sum of the first 10 terms of the series

$\frac{1}{1 + 1 ^{4} \cdot 4} + \frac{2}{1 + 2 ^{4} \cdot 4} + \frac{3}{1 + 3 ^{4} \cdot 4} + \frac{4}{1 + 4 ^{4} \cdot 4} + \dots$

is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to:

A
256
B
264
C
276
D
284

Solution

The general term is $T_{r} = \frac{r}{1 + 4 r ^{4}}$ . Factor the denominator as a difference of squares: $1 + 4 r^{4} = (2 r^{2} + 1)^{2} - (2 r)^{2} = (2 r^{2} + 2 r + 1) (2 r^{2} - 2 r + 1)$ .

Then $T_{r}$ telescopes:

$T_{r} = \frac{1}{4} [\frac{1}{2 r ^{2} - 2 r + 1} - \frac{1}{2 r ^{2} + 2 r + 1}]$ .

Summing the first 10 terms collapses to $\frac{1}{4} [\frac{1}{1} - \frac{1}{221}] = \frac{220}{4 \cdot 221} = \frac{55}{221}$ .

$g cd (55, 221) = 1$ , so $m + n = 55 + 221 = 276$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q5: Summation of Series

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