= Solution
For <nilpotent elements> $a,b$ of the <commutative ring>, choose positive integers $r,s$ with $a^r=b^s=0$. The <binomial theorem> gives
$$
(a+b)^{r+s-1}=\sum_{j=0}^{r+s-1}\binom{r+s-1}{j}a^jb^{r+s-1-j}=0:
$$
each term has either $j\ge r$ or $r+s-1-j\ge s$. Also $(-a)^r=0$, $0\in N$, and $(ta)^r=t^ra^r=0$ for every $t\in R$. Thus $N$ is an additive subgroup stable under multiplication by arbitrary ring elements:
$$
\boxed{N\text{ is an ideal of }R,\text{ its nilradical}.}
$$
Commutativity is essential to the simple binomial argument; an arbitrary sum of nilpotent elements in a noncommutative ring need not be nilpotent.
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