Problem - 3656
Let the domain of function $f(n)$ be $\mathbb{N}$, $f(1)=1$, and for any $m, n\in\mathbb{N}$, $$f(m+n)=f(m)+f(n)+mn$$
Determine $f(n)$.
Let $m=1$, then $f(n+1)=f(1) + f(n)+n=f(n)+ (n + 1)$.
Therefore
\begin{align*}
f(2) &= f(1) + 2\\
f(3) &= f(2) + 3\\
\cdots\\
f(n) &= f(n-1) + n
\end{align*}
Adding these relations gives:
$$f(n) = f(1)+2+3+\cdots + n = 1+2+3+\cdots+n=\frac{n(n+1)}{2}$$