Sum of Ideals is Ideal/General Result

Theorem
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\left({R, +, \circ}\right)$.

Then:
 * $J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.

Proof
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\left({R, +, \circ}\right)$.

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

$P(1)$ is true, as this just says $J_1$ is an ideal of $R$.

Basis for the Induction
$P(2)$ is the case:
 * $J_1 + J_2$ is an ideal of $R$

which is proved in Sum of Ideals is Ideal.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $J_1 + J_2 + \cdots + J_k$ is an ideal of $R$.

Then we need to show:
 * $J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$.

Induction Step
This is our induction step:

Let $J = J_1 + J_2 + \cdots + J_k$.

From the induction hypothesis, $J$ is an ideal.

From the base case, $J + J_{k+1}$ is an ideal.

That is:
 * $J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.