summation - Induction: $sum_{k=1}^{2n} (-1)^k k = n$

Use the proof of induction to show : $\sum_{k=1}^{2n} (-1)^k k = n$

I know how to show the base step of this problem, but in showing the inductive step I am having trouble determining how to show they are equal.

Answer

Suppose, it is true for $n-1$.
Then

$$\sum_{k=1}^{2n} (-1)^k k =$$ $$=\left( \sum_{k=1}^{2(n-1)} (-1)^k k\right)+(-1)^{2n-1}\cdot(2n-1)+(-1)^{2n}\cdot(2n)=$$
$$=(n-1)-(2n-1)+2n=n$$
it is also true for $n$.