Let f_k(x) = 1 k ( ^k x + ^k x) where k N , then f₆(x) - f₄(x) =

Step-by-step solution

Given f_k(x) = 1 k ( ^k x + ^k x) For k = 6 : f₆(x) = 1 6 ( ^6 x + ^6 x) Using the identity a^3 + b^3 = (a+b)^3 - 3ab(a+b) with a = ^2 x and b = ^2 x : ^6 x + ^6 x = ( ^2 x + ^2 x)^3 - 3 ^2 x ^2 x( ^2 x + ^2 x) = 1 - 3 ^2 x ^2 x Thus, f₆(x) = 1 6 (1 - 3 ^2 x ^2 x) = 1 6 - 1 2 ^2 x ^2 x For k = 4 : f₄(x) = 1 4 ( ^4 x + ^4 x) Using the identity a^2 + b^2 = (a+b)^2 - 2ab with a = ^2 x and b = ^2 x : ^4 x + ^4 x = ( ^2 x + ^2 x)^2 - 2 ^2 x ^2 x = 1 - 2 ^2 x ^2 x Thus, f₄(x) = 1 4 (1 - 2 ^2 x ^2 x) = 1 4 - 1 2 ^2 x ^2 x