4 Let $a_{1}, a_{2}, \cdots, a_{2 m+1}$ be $2 m+1$ real numbers, with adjacent numbers $a_{n}, a_{n+1}$ being the roots of the equation $x^{2}+4 n x+b_{n}$ $=0$. Try to find the value of $\sum_{n=1}^{2 m} b_{n}$ when $a_{1}=0$.

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4 From Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a_{n}+a_{n+1}=-4 n \\
a_{n} \cdot a_{n+1}=b_{n}
\end{array}\right.
$$

From (1), we get $\left(a_{n}+2 n-1\right)+\left(a_{n-1}+2(n+1)-1\right)=0$, hence the sequence $\left\{a_{n}+2 n-1\right\}(n=1,2, \cdots, 2 m)$ is a geometric sequence with a common ratio of -1. Therefore,
$$
\begin{array}{l} 
a_{n}+2 n-1=(-1)^{n-1}\left(a_{1}+1\right)=(-1)^{n-1}, \\
a_{n}=-2 n+1+(-1)^{n-1}, \\
a_{n-1}=-2 n-1+(-1)^{n} . \\
b_{n}=a_{n} \cdot a_{n+1}=4 n^{2}-2+2 \cdot(-1)^{n} . \text { Hence } \\
\sum_{n=1}^{2 m} b_{n}=\sum_{n=1}^{2 m}\left(4 n^{2}-2\right)=4 \cdot \frac{2 m(2 m+1)(4 m+1)}{6}-4 m \\
= \frac{8 m}{3}\left(4 m^{2}+3 m-1\right) .
\end{array}
$$