Let for some α ∈ R, f: R → R be a function satisfying f(x+y)=f(x)+2 y^2+y+α x y for all x, y ∈ R. If f(0)=-1 and f(1)=2, then the value of…

Let for some $\displaystyle \alpha \in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $\displaystyle f(x+y)=f(x)+2 y^2+y+\alpha x y$ for all $\displaystyle x, y \in \mathbb{R}$. If $\displaystyle f(0)=-1$ and $\displaystyle f(1)=2$, then the value of $\displaystyle \sum_{n=1}^5(\alpha+f(n))$ is: