Solution
= Solution
The identity $(n+1)^3-n^3=3n(n+1)+1$, summed from zero to $N$, gives $(N+1)^3=3\sum_{n=0}^Nn(n+1)+(N+1)$. Hence
$$
\boxed{\sum_{n=0}^Nn(n+1)=\frac{N(N+1)(N+2)}3.}
$$
= Solution
The identity $(n+1)^3-n^3=3n(n+1)+1$, summed from zero to $N$, gives $(N+1)^3=3\sum_{n=0}^Nn(n+1)+(N+1)$. Hence
$$
\boxed{\sum_{n=0}^Nn(n+1)=\frac{N(N+1)(N+2)}3.}
$$