Given $k \in N$ and a real number $a>0$, under the following conditions,
$$
k_{1}+k_{2}+\cdots+k_{r}=k, k_{i} \in N, 1 \leqslant r \leqslant k .
$$

Find the maximum value of $a^{k_{1}}+a^{k_{2}}+\cdots+a^{k}$.
(Pan Fu supplied the problem)
For $a>0$ and $s, t \in \mathbb{N}$, it is obvious that
$$
a\left(a^{t-1}-1\right)\left(a^{t-1}-1\right) \geqslant 0,
$$

thus we have
$$
a^{t}+a^{t} \leqslant a+a^{t+t-1}.
$$

Using this inequality, we can obtain
$$
a^{k_{1}}+a^{k_{2}}+\cdots+a^{k} \leqslant (r-1) a+a^{k-(r-1)}.
$$

Since the function $a x+a^{k-x}$ is convex, and the maximum value of a convex function must be achieved at the endpoints of the interval, we have
$$
\begin{array}{l}
(r-1) a+a^{k-(r-1)} \\
\leqslant \max \left\{a^{k}, k a\right\}. \quad 1 \leqslant r \leqslant k.
\end{array}
$$

From (1) and (2), we get
$$
a^{k_{1}}+a^{k_{2}}+\cdots+a^{k} \leqslant \max \left\{a^{k}, k a\right\}.
$$

Since both numbers in the right-hand side of the above inequality can be achieved, the maximum value we seek is
$$
\begin{array}{l}
\max \left\{k a, a^{k}\right\} \\
=\left\{\begin{array}{l}
k a, \text { when } a \leqslant k^{\frac{1}{k-1}} \text { and } k \geqslant 2; \\
a^{k}, \text { when } k=1 \text { or } a>k^{\frac{1}{k-1}}, k \geqslant 2.
\end{array}\right.
\end{array}
$$