JEE Main 2025 Apr 2 Shift 2, Mathematics Q22: Summation of Series

JEE Main2025Apr 2, Shift 2Mathematics

If the sum of the first 10 terms of the series $\frac{4 \cdot 1}{1 + 4 \cdot 1 ^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2 ^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3 ^{4}} + \dots$ is $\frac{m}{n}$ , where $g cd (m, n) = 1$ , then $m + n$ is equal to ________.

Your answer

Solution

$T_{r} = \frac{4 r}{1 + 4 r ^{4}}$

$= \frac{4 r}{( 2 r ^{2} + 2 r + 1 ) ( 2 r ^{2} - 2 r + 1 )}$

$= \frac{( 2 r ^{2} + 2 r + 1 ) - ( 2 r ^{2} - 2 r + 1 )}{( 2 r ^{2} + 2 r + 1 ) ( 2 r ^{2} - 2 r + 1 )}$

$= \frac{1}{2 r ^{2} - 2 r + 1} - \frac{1}{2 r ^{2} + 2 r + 1}$

$T_{1} = \frac{1}{1} - \frac{1}{5}$

$T_{2} = \frac{1}{5} - \frac{1}{13}$

$⋮$

$T_{10} = \frac{1}{181} - \frac{1}{221}$

$∴ S_{10} = 1 - \frac{1}{221} = \frac{220}{221} = \frac{m}{n}$

$m + n = 220 + 221 = 441$

$441$

Concept behind this question

Summation of SeriesNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 2 Shift 2, Mathematics Q22: Summation of Series

Concept behind this question

More previous year questions on Summation of Series

Practice more Mathematics