If the sum of the first 10 terms of the series (4 · 1)/(1+4 · 1⁴)+(4 · 2)/(1+4 · 2⁴)+(4 · 3)/(1+4 · 3⁴)+… .. is m/n, where gcd(m, n)=1, then m+n is equal to _______________

Solution

Maths sir · 4 steps

Factor 1+4r⁴ as a difference of squares, and the terms cancel in a chain.

1. Factor the denominator. Add and subtract 4r²: 1+4r⁴ = (1+2r²)²-(2r)² = (2r²+2r+1)(2r²-2r+1).
2. Split the term. The two factors differ by exactly 4r, which is the numerator. So the general term is the difference of their reciprocals.

   4r/(1+4r⁴) = 1/(2r²-2r+1) - 1/(2r²+2r+1)

3. Watch it telescope. Put r+1 in place of r in 2r²-2r+1 and you get 2r²+2r+1. So the second fraction of each term cancels the first fraction of the next term. Only the first fraction at r=1 and the second fraction at r=10 survive.
4. So. The sum is 1/1 - 1/(200+20+1) = 1-1/221 = 220/221. Here 220 = 2² × 5 × 11 and 221 = 13 × 17 share no factor, so m+n = 220+221 = 441.

The common slip: 361. Using 2r²-2r+1 at r=10, which is 181, as the last surviving denominator gives 180/181 and a total of 361, not 441.

The idea to keep. Whenever you see 1+4r⁴ or r⁴+r²+1 below the line, complete the square to factor it. If the numerator equals the difference of the two factors, the series telescopes.

Answer from the NTA final answer key. Solution by PrepWiser.