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 11+14⋅4+21+24⋅4+31+34⋅4+⋯\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to:
A256256
B264264
C276276correct
D284284
Solution
Step 1: Identify the general term
Tr=r1+4r4T_r = \frac{r}{1+4r^4}
 Step 2: Factor the denominator using Sophie Germain identity
1+4r4=(2r2+2r+1)(2r2−2r+1)1+4r^4 = (2r^2+2r+1)(2r^2-2r+1)
 Step 3: Decompose into a telescoping difference Since (2r2+2r+1)−(2r2−2r+1)=4r(2r^2+2r+1)-(2r^2-2r+1)=4r:
Tr=r(2r2+2r+1)(2r2−2r+1)=14(12r2−2r+1−12r2+2r+1)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 2r2+2r+1∣r=k=2(k+1)2−2(k+1)+1∣2r^2+2r+1\big|_{r=k} = 2(k+1)^2-2(k+1)+1\big|, confirming the telescope. Therefore:
S10=14(12(1)2−2(1)+1−12(10)2+2(10)+1)=14(1−1221)=14⋅220221=55221S_{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⁡\gcd and compute m+nm+n Since 221=13×17221=13\times17 and 55=5×1155=5\times11, gcd⁡(55,221)=1\gcd(55,221)=1. Hence m=55m=55, n=221n=221:
m+n=276m+n = 276
 Answer: (3)
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