13. If the expansion of $(a+b)^{n}$ has three consecutive terms whose binomial coefficients form an arithmetic sequence, then the largest three-digit positive integer $n=$
$$
\begin{array}{l}
2 C_{n}^{k+1}=C_{n}^{k}+C_{n}^{k+2} \Rightarrow \frac{2 n!}{(k+1)!(n-k-1)!}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+2)!(n-k-2)!} \\
\Rightarrow 2(k+2)(n-k)=(k+1)(k+2)+(n-k-1)(n-k) \\
\Rightarrow n^{2}-(4 k+5) n+4 k^{2}+8 k+2=0 \Rightarrow n=\frac{4 k+5 \pm \sqrt{8 k+17}}{2} .
\end{array}
$$

Obviously, if $n$ is an integer, $8 k+17$ is an odd perfect square. Let $8 k+17=(2 m+1)^{2}$, then $n=\frac{2 m^{2}+2 m-3 \pm(2 m+1)}{2} \Rightarrow n=(m+1)^{2}-2$ or $m^{2}-2$.
Therefore, the largest three-digit positive integer $n=31^{2}-2=959$.