3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the first term $a$ and common difference $d$. Define
$$
\begin{array}{l}
S_{n}^{(0)}=a_{n}, \\
S_{n}^{(k)}=S_{1}^{(k-1)}+S_{2}^{(k-1)}+\cdots+S_{n}^{(k-1)}(k>0) .
\end{array}
$$

Then, $S_{n}^{(m)}=$ $\qquad$ (express in terms of $a$, $d$, and $n$).
3. $\mathrm{C}_{n+m-1}^{m} a_{1}+\mathrm{C}_{n+m-1}^{m+1} d$.

Notice that
$$
\begin{array}{l}
S_{n}^{(1)}=a_{1}+a_{2}+\cdots+a_{n}=\mathrm{C}_{n}^{1} a_{1}+\mathrm{C}_{n}^{2} d, \\
S_{n}^{(2)}=S_{1}^{(1)}+S_{2}^{(1)}+\cdots+S_{n}^{(1)} \\
=\left(\mathrm{C}_{1}^{1}+\mathrm{C}_{2}^{1}+\cdots+\mathrm{C}_{n}^{1}\right) a_{1}+ \\
\quad\left(\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n}^{2}\right) d \\
=\mathrm{C}_{n+1}^{2} a_{1}+\mathrm{C}_{n+1}^{3} d .
\end{array}
$$

Using mathematical induction, we can prove:
$$
S_{n}^{(m)}=\mathrm{C}_{n+m-1}^{m} a_{1}+\mathrm{C}_{n+m-1}^{m+1} d .
$$