JEE Main 2025 Apr 8 Shift 2, Mathematics Q5: Summation of Series

JEE Main2025Apr 8, Shift 2Mathematics

If $\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots = \frac{π ^{4}}{90}$ , $\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots = α$ , $\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots = β$ , then $\frac{α}{β}$ is equal to

A
$23$
B
$18$
C
$15$
D
$14$

Solution

Factor the even-index sum: $β = k = 1 \sum \infty \frac{1}{( 2 k ) ^{4}} = \frac{1}{16} k = 1 \sum \infty \frac{1}{k ^{4}} = \frac{1}{16} \cdot \frac{π ^{4}}{90} = \frac{π ^{4}}{1440}$ .

The odd-index sum is the total minus the even-index sum: $α = \frac{π ^{4}}{90} - \frac{π ^{4}}{1440} = \frac{16 π ^{4} - π ^{4}}{1440} = \frac{15 π ^{4}}{1440} = \frac{π ^{4}}{96}$ .

Taking the ratio: $\frac{α}{β} = \frac{π ^{4} /96}{π ^{4} /1440} = \frac{1440}{96} = 15$ , matching option (3).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q5: Summation of Series

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