Proving by Contrapositive I am given this statement ... Use a proof by contrapositive to show that if n is an integer and n^2 is odd, then n
must be odd.
Since its prove by contrapositive I have to to assume the negation. which is Assuming n is an even integer and that n^2 is even as well. By definition n could be represented as 2k (2 for  some k). To proof the statement do i just plug in 2k and square it to show that by definition it is even and not odd?
 A: The contrapositive to a statement "$P$ implies $Q$" is "not $Q$ implies not $P$". Symbolically,
$$(P \implies Q) \iff (\neg Q \implies \neg P)$$
That is, you not only assume the negation but you effectively swap around what implies what.
Thus, if you are trying to prove "if $n$ is an integer with $n^2$ odd, then $n$ is odd", then the contrapositive is "if $n$ is an even integer, then $n^2$ is even." This is basically what you said, but notice how it is a little more nuanced than simply negating the statement. (Above, $P$ can be taken as "$n^2$ is odd" and $Q$ as "$n$ is odd.")
To prove this statement, you can thus indeed say $n=2k$ for some integer $k$, and then show $2$ divides $n^2$, making it even. This follows quite easily, as...

 $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$

A: Yep.  This is a relatively easy one and you figured it out.

$n^2$ is odd $\implies n$ is odd

is equivalent to

$n$ is even $\implies n^2$ is even.

And we prove that:
If $n$ is even then there is an integer $k$ so that $n = 2k$.
Then $(2k)^2 = 4k^2 = 2(2k^2)$.  And as $2k^2$ is an integer $2(2k^2)=n^2$ is even.
That's it.  We're done.
A: For contrapositive proof you have to assume that $n$ is even and prove that $n^2$ is even. 
In general $P\implies Q$ is equivalent to  $\lnot Q\implies\lnot P$
and that is why you need to switch them around and make negations.
