Prove even integer sum using induction This is a homework problem, so please do not give the answer away. I must prove the following using mathematical induction:

$\forall n\in\mathbb{Z^+},\;2+4+6+\cdots+2n=n^2+n.$

This is what I have so far:

Let $P\left(n\right)$ represent $2+4+6+\cdots+2n=n^2+n$. Since
  $P\left(1\right)=\left(1\right)^2+1=2$, $P\left(1\right)$ is true. If
  $P\left(k\right)$ is true for $k\in\mathbb{Z^+}$, then
  $P\left(k+1\right)$ is true. Hence \begin{align} n^2+n & =
 \left(k+1\right)^2+\left(k+1\right)\notag\\ & = k^2 + 2k + 1 + k +
 1\notag\\ & = k^2+3k + 2\notag\\
 &=\left(k+2\right)\left(k+1\right).\notag \end{align}

I feel the last few steps do not do justice. Could you provide a hint to get me back on track? Thank you for your time,

Suppose $P(k)$ is true for some $k \in \mathbb{N}$. That is, 
\begin{equation}
2+4+6+\cdots+2k=k^2+k.
\end{equation}
Now, we need to use this to show
\begin{equation}
2+4+6+\cdots+2k+2(k+1)=(k+1)^2+(k+1).
\end{equation}
I hope this helps. 
