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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · medium
Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0, where p,q∈Zp, q \in \mathbb{Z}. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in G.P., then ∣p+q∣|p + q| equals:
Sequences & Series · easy
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
Sequences & Series · medium
Let a1,a2,…,a4001a_1, a_2, \ldots, a_{4001} be in A.P. such that 1a1a2+1a2a3+⋯+1a4000a4001=10\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10 and a2+a4000=50a_2+a_{4000}=50. Then ∣a1−a4001∣|a_1-a_{4001}| equals
Sequences & Series · medium
If p+q⋅5xp−q⋅5x=q+r⋅5xq−r⋅5x=r+s⋅5xr−s⋅5x\frac{p+q\cdot 5^x}{p-q\cdot 5^x}=\frac{q+r\cdot 5^x}{q-r\cdot 5^x}=\frac{r+s\cdot 5^x}{r-s\cdot 5^x}, then p,q,r,sp, q, r, s are in

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath