10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=1+\frac{1}{n} \sum_{k=1}^{n} a_{k} \text {. }
$$
(1) Write out the first seven terms of the sequence;
(2) For any positive integer $n$, find the expression for $a_{n}$.
10. (1) Calculate in sequence
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=\frac{5}{2}, a_{4}=\frac{17}{6}, \\
a_{5}=\frac{37}{12}, a_{6}=\frac{197}{60}, a_{7}=\frac{207}{60} .
\end{array}
$$
(2) Notice that,
$$
\begin{array}{l}
a_{n+1}=1+\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}, \\
a_{n}=1+\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1} .
\end{array}
$$

Then $a_{n+1}-a_{n}$
$$
\begin{aligned}
= & \frac{a_{1}+a_{2}+\cdots+a_{n}}{n}-\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1} \\
= & \frac{a_{n}}{n}+\left(\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n}-\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}\right) \\
= & \frac{1}{n}\left(1+\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}\right)+ \\
& \left(a_{1}+a_{2}+\cdots+a_{n-1}\right)\left(\frac{1}{n}-\frac{1}{n-1}\right) \\
= & \frac{1}{n} .
\end{aligned}
$$

Therefore, $a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right)$
$$
\begin{array}{l}
\quad=1+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}\right) \\
=2+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1} .
\end{array}
$$