Solution
= Solution
Change the summation index to $j=n+1$. Then $j$ runs from $-99$ to $100$. Each term $j^3$ for $1\leq j\leq99$ cancels its partner $(-j)^3$, and the zero term contributes nothing. The sole unpaired term is $100^3$, so
$$
\boxed{\sum_{n=-100}^{99}(n+1)^3=10^6.}
$$