<s>[INST] Use Proof of Induction to prove $\sum_{k=1}^{2n} (-1)^k k = n$ Base Case:
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n\\
(-1)^1 (1) + (-1)^2(2) &=&1 \\
1=1 
\end{eqnarray*}
Inductive Step: For this step we must prove that 
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n \Rightarrow  \sum_{k=1}^{2(n+1)} (-1)^k k = n+1 \\
\sum_{k=1}^{2(n+1)} (-1)^k k = \sum_{k=1}^{2n} (-1)^k k + \sum_{k=1}^{2} (-1)^k k
\end{eqnarray*}
We know that, 
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n 
\end{eqnarray*}
and from the base case we conclude, 
\begin{eqnarray*}
\sum_{k=1}^{2} (-1)^k k = 1
\end{eqnarray*}
Therefore in we have $n+1 = n+1$. Is this a method that could be utilized?
 [/INST] Basically, yes. There is a bit of a snag with your second summation though: it should be $k = 2n + 1$ and $k = 2n + 2$. So we have:
\begin{align*}
\sum_{k=1}^{2(n + 1)} (-1)^k k
&= \sum_{k=1}^{2n} (-1)^k k + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) \\
&= n + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) &\text{by the induction hypothesis}\\
&= n - (2n + 1) + (2n + 2)\\
&= n + 1\\
\end{align*}
as desired.
</s>