Sum of Even Integers is Even

Theorem
The sum of any finite number of even integers is itself even.

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * The sum of $n$ even integers is an even integer.

$P(1)$ is trivially true, as this just says:
 * The sum of $1$ even integers is an even integer.

The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even.

So $P(0)$ is also true.

Basis for the Induction
$P(2)$ is the case:
 * The sum of any two even integers is itself even

which has been proved in Sum of Even Integers is Even.

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:
 * The sum of any $k$ even integers is itself even.

Then we need to show:
 * The sum of any $k+1$ even integers is itself even.

Induction Step
This is our induction step:

Consider the sum of any $k+1$ even integers.

This is the sum of $k$ even integers (which is even by the induction hypothesis) and another even integer.

That is, it is the sum of two even integers, and therefore, by the basis for the induction, also even.

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

That is:
 * The sum of any finite number of even integers is itself even.