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← JEE Main 2026 · 5 Apr · Shift 2
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JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 5

If the sum of the first 10 terms of the series 11+14×4+21+24×4+31+34×4+41+44×4+…\dfrac1{1+1^4\times4}+\dfrac2{1+2^4\times4}+\dfrac3{1+3^4\times4}+\dfrac4{1+4^4\times4}+\dots is mn\dfrac mn, gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to :

Choose one option
Answer and worked solution

Official answer: C

Factor. 1+4n4=(2n2+2n+1)(2n2−2n+1)1+4n^4=(2n^2+2n+1)(2n^2-2n+1), so
n1+4n4=14[12n2−2n+1−12n2+2n+1].\frac n{1+4n^4}=\frac14\left[\frac1{2n^2-2n+1}-\frac1{2n^2+2n+1}\right].

Telescoping. 2(n+1)2−2(n+1)+1=2n2+2n+12(n+1)^2-2(n+1)+1=2n^2+2n+1, so the sum of the first 10 terms is
14(1−1221)=55221.\frac14\left(1-\frac1{221}\right)=\frac{55}{221}.

Answer. m+n=276m+n=276.

(C)

Topics:Special series sums
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