Let f: N N be a function such that f(x+y) = f(x) + f(y) + 2xy - 1 for all x, y N . If f(1) = 2 , then the value of _ k=1 ¹⁵ f(k) is

Step-by-step solution

Given the functional equation f(x+y) = f(x) + f(y) + 2xy - 1 . Substituting y = 1 , we get: f(x+1) = f(x) + f(1) + 2x(1) - 1 Since f(1) = 2 , this becomes: f(x+1) = f(x) + 2 + 2x - 1 f(x+1) - f(x) = 2x + 1 To find f(n) , we can sum this difference equation from x = 1 to n-1 : _ x=1 ^ n-1 (f(x+1) - f(x)) = _ x=1 ^ n-1 (2x + 1) The left side telescopes to f(n) - f(1) . The right side is an arithmetic progression: f(n) - 2 = 2 ( (n-1)n 2 ) + (n-1) f(n) - 2 = n^2 - n + n - 1 f(n) = n^2 + 1 We need to evaluate _ k=1 ¹⁵