4・133 Let $x_{1}, x_{2}, \cdots, x_{7}$ be real numbers satisfying the following equations:
$$\begin{array}{l}
\sum_{k=1}^{7} k^{2} x_{k}=1 \\
\sum_{k=1}^{7}(k+1)^{2} x_{k}=12 \\
\sum_{k=1}^{7}(k+2)^{2} x_{k}=123
\end{array}$$

Find the value of $\sum_{k=1}^{7}(k+3)^{2} x_{k}$.
[Solution] Let $a \cdot k^{2}+b(k+1)^{2}+c \cdot(k+2)^{2}=(k+3)^{2}$. By equating the corresponding coefficients, we get
$$\left\{\begin{array}{l}
a+b+c=1 \\
2 b+4 c=6 \\
b+4 c=9
\end{array}\right.$$

Solving these equations, we get $a=1, b=-3, c=3$.
Thus, (1) + (2) $\times(-3)+(3) \times 3$ gives
$$\sum_{k=1}^{7}(k+3)^{2} x_{k}=1-3 \times 12+3 \times 123=334$$