Example 7 Let $x_{1}, x_{2}, \cdots, x_{7}$ all be integers, and
$$
\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6} \\
+49 x_{7}=1, \\
4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6} \\
+64 x_{7}=12, \\
9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6} \\
+81 x_{7}=123 .
\end{array}
$$

Find $16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+$
$$
81 x_{6}+100 x_{7}=
$$
. $\qquad$
Solution: Since the coefficients of the same letters in the four equations are the squares of four consecutive natural numbers, i.e., $n^{2}, (n+1)^{2}, (n+2)^{2}, (n+3)^{2}$, to find the value of (4), it is necessary to express (4) using (1), (2), and (3), i.e., to express $(n+3)^{2}$ using $n^{2}, (n+1)^{2}, (n+2)^{2}$. Therefore, we can set
$$
(n+3)^{2}=a(n+2)^{2}+b(n+1)^{2}+c^{2} \text {. }
$$
$$
\text { Let } n=0 \text {, we get } 9=4 a+b \text {; }
$$

Taking $n:=-1$, we get $4=a \div c$;
Taking $n:=\cdots$, we get $1=b+4 c$.
Solving, we get $a=3, b=-3, c=1$.
Thus, $(n+3)^{2}=3(n+2)^{2}-3(n+1)^{2}+n^{2}$.
Therefore, $16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+$ $81 x_{6}+100 x_{7}$
$$
=3 \times 123-3 \times 12+1=334 \text {. }
$$