If the sum of the first 20 terms of the series (4 · 1)/(4+3 · 1²+1⁴)+(4 · 2)/(4+3 · 2²+2⁴)+(4 · 3)/(4+3 · 3²+3⁴)+(4 · 4)/(4+3 · 4²+4⁴)+… · is m/n, where m and n are coprime, then m+n is equal to :

Solution

Maths sir · 5 steps

Write the denominator as a difference of squares, and the terms cancel down the line.

1. Factor the denominator. The general term is 4r/(r⁴ + 3r² + 4). Write r⁴ + 3r² + 4 = (r² + 2)² - r² = (r² - r + 2)(r² + r + 2).
2. Split the term. The two factors differ by exactly 2r, and the numerator 4r is twice that. So each term is twice the difference of two reciprocals.

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

3. Telescope. Put r + 1 in place of r in r² - r + 2 and you get r² + r + 2. So the negative part of each term cancels the positive part of the next one. Only the first part of term 1 and the last part of term 20 survive.
4. Add up. The sum is 2(1/2 - 1/(20² + 20 + 2)) = 2(1/2 - 1/422) = 1 - 1/211 = 210/211.
5. So. m = 210 and n = 211 are coprime, so m + n = 421, which is option (b).

The idea to keep. When you see r⁴ + ar² + b², write it as (r² + b)² - (something)². If the numerator is a multiple of the gap between the two factors, the series telescopes.

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