Let for some R , f: R R be a function satisfying f(x+y)=f(x)+2y^2+y+ xy for all x,y R . If f(0)=-1 and f(1)=2 , then the value of _ n=1 ⁵( +f(n)) is:

Step-by-step solution

Given f(x+y) = f(x) + 2y^2 + y + xy Substituting x = 0 and replacing y with x , we get: f(x) = f(0) + 2x^2 + x + (0)x Since f(0) = -1 , we have: f(x) = 2x^2 + x - 1 To find , substitute f(x) into the original functional equation: 2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + xy 2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + xy Comparing both sides, we get = 4 . We need to find the value of _ n=1 ⁵( +f(n)) : _ n=1 ⁵(4 + 2n^2 + n - 1) = _ n=1 ⁵(2n^2 + n + 3) = 2 _ n=1 ⁵n^2 + _ n=1 ⁵n + _ n=1 ⁵3 = 2 (