[proofplan] We prove the result by induction on $n$. The base case $n = 1$ is immediate. For the inductive step, we write $M_1 \oplus \cdots \oplus M_{k+1}$ as an extension of $M_1 \oplus \cdots \oplus M_k$ by $M_{k+1}$ via a canonical short exact sequence, and then apply the [Noetherian Property in Short Exact Sequences](/theorems/2901). The artinian case follows by the same argument, since the cited theorem covers both chain conditions. [/proofplan] [step:Verify the base case $n = 1$] When $n = 1$, the direct sum $M_1 \oplus \cdots \oplus M_n = M_1$ is noetherian (resp. artinian) by hypothesis. [/step] [step:Establish the short exact sequence for the inductive step] For $k \geq 1$, consider the inclusion and projection maps \begin{align*} i: M_1 \oplus \cdots \oplus M_k &\to M_1 \oplus \cdots \oplus M_k \oplus M_{k+1}, \quad (m_1, \ldots, m_k) \mapsto (m_1, \ldots, m_k, 0), \\ \pi: M_1 \oplus \cdots \oplus M_k \oplus M_{k+1} &\to M_{k+1}, \quad (m_1, \ldots, m_k, m_{k+1}) \mapsto m_{k+1}. \end{align*} The map $i$ is injective, $\pi$ is surjective, and $\ker \pi = \operatorname{im} i$ (an element maps to $0$ under $\pi$ if and only if its last component is $0$, which is precisely the image of $i$). Therefore the sequence \begin{align*} 0 \to M_1 \oplus \cdots \oplus M_k \xrightarrow{i} M_1 \oplus \cdots \oplus M_{k+1} \xrightarrow{\pi} M_{k+1} \to 0 \end{align*} is a short exact sequence of $R$-modules. [/step] [step:Complete the induction using the noetherian property in short exact sequences] Suppose the result holds for $k$ modules, i.e., $M_1 \oplus \cdots \oplus M_k$ is noetherian (resp. artinian). By hypothesis, $M_{k+1}$ is also noetherian (resp. artinian). The short exact sequence from the previous step has both end terms noetherian (resp. artinian). By the [Noetherian Property in Short Exact Sequences](/theorems/2901), the middle term $M_1 \oplus \cdots \oplus M_{k+1}$ is noetherian (resp. artinian). By induction, the result holds for all $n \geq 1$. [/step]