Given the functional equation f(x+y)=f(x)+f(y)+xyf(x+y) = f(x) + f(y) + xyf(x+y)=f(x)+f(y)+xy, where f(1)=2f(1) = 2f(1)=2, find f(n)f(n)f(n) for n∈Z+n \in \mathbb{Z}^+n∈Z+.
f(n)=n2+nf(n) = n^2 + nf(n)=n2+n
f(n)=n(n+1)2f(n) = \frac{n(n+1)}{2}f(n)=2n(n+1)
f(n)=2nf(n) = 2nf(n)=2n
f(n)=n2f(n) = n^2f(n)=n2