If the sum of the first ten terms of the series
1/5 + 2/65 + 3/325 + 4/1025 + 5/2501 + ....
is m/n, where m and n are co-prime numbers, then m + n is equal to ______________.

Solution

Maths sir · 4 steps

Spot that each denominator is 4r⁴+1, factor it as a difference of squares, and the sum telescopes.

1. Find the general term. The denominators 5, 65, 325, 1025, 2501 are 4r⁴+1 for r = 1, 2, 3, 4, 5. So the r-th term is Tᵣ = r/(4r⁴+1).
2. Factor the denominator. Write 4r⁴+1 = (2r²+1)²-(2r)² = (2r²-2r+1)(2r²+2r+1). The two factors differ by exactly 4r, so splitting into partial fractions gives the identity below.

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

3. Telescope. The second fraction for r equals the first fraction for r+1, because 2(r+1)²-2(r+1)+1 = 2r²+2r+1. Adding from r=1 to 10, everything cancels except the first and last pieces: S₁₀ = 1/4(1-1/221).
4. So. S₁₀ = 1/4×220/221 = 55/221. Since 55 = 5× 11 and 221 = 13× 17, they are co-prime. So m+n = 55+221 = 276.

The common slip: 441. Dropping the factor 1/4 in front gives 220/221 and so m+n = 441.

The idea to keep. When the denominator is 4r⁴+1 or r⁴+r²+1, write it as a difference of squares. The two factors usually differ by a multiple of r, and the sum telescopes.

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