Example 3 (1990 Japan First Round Selection Contest) Let $a_{n}=50+n^{2}, n=1,2, \cdots$ Find the maximum value of the greatest common divisor $d_{n}=\left(a_{n}, a_{n+1}\right)$ of $a_{n}$ and $a_{n+1}$.
Solve for $d$:
$$
\begin{aligned}
d_{n} & =\left(a_{n}, a_{n+1}\right)=\left(50+n^{2}, 50+(n+1)^{2}-\left(50+n^{2}\right)\right) \\
& =\left(50+n^{2}, 2 n+1\right)=\left(2\left(50+n^{2}\right), 2 n+1\right) \text{ (since } 2 n+1 \text{ is odd }) \\
& =\left(2\left(50+n^{2}\right)-n(2 n+1), 2 n+1\right)=(100-n, 2 n+1) \\
& =(100-n, 2 n+1+2(100-n))=(100-n, 201) \leqslant 201 .
\end{aligned}
$$

When $n=100 \neq 201 k\left(k \in \mathbf{N}_{+}\right)$, $d_{n}=201$, so the maximum value of $\left(a_{n}, a_{n+1}\right)$ is 201.