8.7 Given $a_{1}=1996$,
$$
a_{k}=\left[\sqrt{a_{1}+a_{2}+\cdots+a_{k-1}}\right], k=2,3, \cdots
$$

where $[x]$ denotes the greatest integer not exceeding $x$. Find $a_{1966}$.
[Solution] $a_{1}=1966=44^{2}+30$, so $a_{2}=a_{3}=44$ and
$$
a_{1}+a_{2}+a_{3}=45^{2}+29 .
$$

From this, we know that $a_{4}=a_{5}=45$ and
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=46^{2}+28 \text {. }
$$

By analogy, we get $a_{60}=a_{61}=73$ and
$$
a_{1}+a_{2}+\cdots+a_{61}=74^{2} \text {. }
$$

Thus,
$$
\begin{array}{l}
a_{62}=a_{63}=a_{64}=74 \text { and } \\
a_{1}+a_{2}+\cdots+a_{64}=75^{2}+73 .
\end{array}
$$

Repeating the previous proof, we get $a_{65}=a_{66}=75, \cdots, a_{209}=a_{210}=147$, $a_{211}=a_{212}=a_{213}=148$ and
$$
a_{1}+a_{2}+\cdots+a_{213}=149^{2}+147 .
$$

In general, it is easy to prove by induction that the sequence $a_{1}, a_{2}, a_{3}, \cdots$ is non-decreasing from the 2nd term onwards, taking all natural numbers not less than 44. Furthermore, numbers of the form $2^{n} \cdot 74$ are taken three times consecutively $(n$ is a non-negative integer), while all other numbers are taken twice consecutively. From this, it is not difficult to calculate
$$
a_{1966}=1024 \text {. }
$$