Sequences & SeriesmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Sequences & Series: Sum First Terms Series Equal (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

If the sum of the first 10 terms of the series

\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots

\dfrac{m}{n}

with

\gcd(m,n)=1

, then

m+n

is equal to:

256

264

276

correct

284

Solution

Step 1: Identify the general term

T_r = \frac{r}{1+4r^4}

Step 2: Factor the denominator using Sophie Germain identity

1+4r^4 = (2r^2+2r+1)(2r^2-2r+1)

Step 3: Decompose into a telescoping difference Since

(2r^2+2r+1)-(2r^2-2r+1)=4r

T_r = \frac{r}{(2r^2+2r+1)(2r^2-2r+1)} = \frac{1}{4}\left(\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}\right)

Step 4: Evaluate the telescoping sum Observe that

2r^2+2r+1\big|_{r=k} = 2(k+1)^2-2(k+1)+1\big|

, confirming the telescope. Therefore:

S_{10} = \frac{1}{4}\left(\frac{1}{2(1)^2-2(1)+1}-\frac{1}{2(10)^2+2(10)+1}\right) = \frac{1}{4}\left(1-\frac{1}{221}\right) = \frac{1}{4}\cdot\frac{220}{221} = \frac{55}{221}

Step 5: Verify

\gcd

and compute

m+n

Since

221=13\times17

and

55=5\times11

\gcd(55,221)=1

. Hence

m=55

n=221

m+n = 276

Answer: (3)

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