Question 2: Find the sum of the terms with integer powers of $x$ in the expansion of $(\sqrt[m]{x}+a)^{n}$, where $m, n \in \mathbf{N}, m, n \geqslant 2$, $a \in \mathbf{R}, a \neq 0$.
When $m=a=2$, it is the original problem.
Let the required sum be $S$.
Obviously, when $m>n$, $S=a^{n}$.
When $m \leqslant n$, from the general term of the expansion of $(a+\sqrt[m]{x})^{n}$, $T_{k+1}=\mathrm{C}_{n}^{k} a^{n-k} x^{\frac{k}{m}}$, we know $S=\sum_{m \mid k} \mathrm{C}_{n}^{k} a^{n-k}$.
First, find the sum $S^{\prime}=\sum_{m \mid j} \mathrm{C}_{n}^{j} q^{j}$.
Let $\varepsilon=\mathrm{e}^{\frac{2 \pi i}{m}}$. Then
$$
\begin{array}{l}
\sum_{k=0}^{m-1}\left(1+q \varepsilon^{k}\right)^{n}=\sum_{j=0}^{n}\left(\mathrm{C}_{n}^{j} q^{j} \sum_{k=0}^{m-1} \varepsilon^{k j}\right) \\
=\sum_{m 1 j} \mathrm{C}_{n}^{j} q^{j}\left(\sum_{k=0}^{m-1} \varepsilon^{k j}\right)=m S^{\prime} \\
\Rightarrow \sum_{m \mid j} \mathrm{C}_{n}^{j} q^{j}=\frac{1}{m} \sum_{k=0}^{m-1}\left(1+q \varepsilon^{k}\right)^{n} .
\end{array}
$$

Therefore, $S=a^{n} \sum_{m \mid k} \mathrm{C}_{n}^{k}\left(\frac{1}{a}\right)^{k}$
$$
\begin{array}{l}
=a^{n} \cdot \frac{1}{m} \sum_{k=0}^{m-1}\left(1+\frac{1}{a} \varepsilon^{k}\right)^{n} \\
=\frac{1}{m} \sum_{k=0}^{m-1}\left(a+\varepsilon^{k}\right)^{n} .
\end{array}
$$

【Note】A more general form of equation (1) is
$$
\sum_{k=r(\bmod m)} \mathrm{C}_{n}^{k} q^{k}=\frac{1}{m} \sum_{k=0}^{m-1} \varepsilon^{-k}\left(1+q \varepsilon^{k}\right)^{n},
$$

where $m \geqslant r \geqslant 0, m \geqslant 2$, and from this we get
$$
\sum_{k=r(\bmod m)} \mathrm{C}_{n}^{k}=\frac{1}{m} \sum_{k=0}^{m-1}\left(2 \cos \frac{k \pi}{m}\right)^{n} \cos \frac{(n-2 r) k \pi}{m} .
$$