If the sum of the first 10 terms of the series 1 1 + 1^4 4 + 2 1 + 2^4 4 + 3 1 + 3^4 4 + 4 1 + 4^4 4 + is m n , (m, n) = 1 , then m + n is equal to :

Step-by-step solution

The general term of the series is T_r = r 1 + 4r^4 . Factorizing the denominator: 1 + 4r^4 = (1 + 2r^2)^2 - 4r^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1) Thus, T_r = r (2r^2 - 2r + 1)(2r^2 + 2r + 1) Multiplying and dividing by 4 : T_r = 1 4 [ 4r (2r^2 - 2r + 1)(2r^2 + 2r + 1) ] T_r = 1 4 [ 1 2r^2 - 2r + 1 - 1 2r^2 + 2r + 1 ] The sum of the first 10 terms is: S₁₀ = _ r=1 ¹⁰ T_r = 1 4 _ r=1 ¹⁰ ( 1 2r^2 - 2r + 1 - 1 2(r+1)^2 - 2(r+1) + 1 ) This is a telescoping series, so all intermediate terms cancel out: S₁₀ = 1 4 [ ( 1 1 -