Example 3 The second-order difference sequence of the sequence $\left\{a_{n}\right\}$ is all 16, and $a_{63}=a_{89}=10$, find $a_{51}$
Let $\left\{a_{n}\right\}$ be a sequence whose first-order difference sequence is $\left\{b_{n}\right\}$.  
Then $\left\{b_{n}\right\}$ is an arithmetic sequence with a common difference of 16. Suppose its first term is $a$, then $b_{n}=a+16(n-1)$. Therefore,  
\[ a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right)=a_{1}+\sum_{k=1}^{n-1} b_{k}=a_{1}+(n-1) a+8(n-1)(n-2) \]  
This is a quadratic polynomial in $n$, with a quadratic coefficient of 8, and $a_{63}=a_{89}=10$. Hence,  
\[ a_{n}=8(n-63)(n-89)+10 \]  
Thus,  
\[ a_{51}=8(51-63)(51-89)+10=3658 \]