JEE Main 2025 Apr 4 Shift 2, Mathematics Q12: Summation of Series

JEE Main2025Apr 4, Shift 2Mathematics

If the sum of the first 20 terms of the series

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

is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to :-

A
423
B
420
C
421
D
422

Solution

Factor the denominator: $r^{4} + 3 r^{2} + 4 = (r^{2} + r + 2) (r^{2} - r + 2)$ . So the general term is

$T_{r} = \frac{4 r}{( r ^{2} + r + 2 ) ( r ^{2} - r + 2 )}$ .

Notice that $(r^{2} + r + 2) - (r^{2} - r + 2) = 2 r$ , so

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

This telescopes because $r^{2} + r + 2 = (r + 1)^{2} - (r + 1) + 2$ . Hence

$r = 1 \sum 20 T_{r} = 2 [\frac{1}{1 - 1 + 2} - \frac{1}{2 0 ^{2} + 20 + 2}] = 2 [\frac{1}{2} - \frac{1}{422}] = \frac{420}{422} = \frac{210}{211}$ .

So $m = 210, n = 211$ , and $m + n = 421$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q12: Summation of Series

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