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,
 A: 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. 
A: A little late to the party here, but I thought I might weigh in to share something useful I learned about summation/product identities when I first learned about induction, especially since this question has the proof-writing tag (I'm only going to focus on the inductive step, as you have already established the base case). 
For most basic problems, the idea is to "peel off the $k+1$th summand" and then use the inductive hypothesis appropriately, etc. I'll give you an example concerning your problem specifically (since you have already accepted an answer, I imagine you have already proved the result). 
You are trying to show that 
$$
2+4+6+\cdots+2n=n^2+n\tag{1}
$$ for all $n\geq 1$, where $n\in\mathbb{Z^+}$. Notice that we can represent $(1)$ by using $\Sigma$-notation:
$$
\sum_{i=1}^n 2i=n^2+n.\tag{2}
$$
Thus, we are really trying to prove that $(2)$ holds for all $n\in\mathbb{Z^+}$. To this end, let $P(n)$ denote the statement
$$
P(n) : \sum_{i=1}^n 2i=n^2+n.
$$
Fix some $k\geq 1$ and assume $P(k)$ to be true; that is,
$$
P(k) : \sum_{i=1}^k 2i=k^2+k
$$
holds. To be shown is that $P(k+1)$, where
$$
P(k+1) : \sum_{i=1}^{k+1} 2i=(k+1)^2+(k+1),
$$
follows. Beginning with the left-hand side of $P(k+1)$,
\begin{align}
\sum_{i=1}^{k+1} 2i &= \underbrace{\sum_{i=1}^k 2i + 2(k+1)}_{\text{"peel off the $k+1$th summand"}}\tag{by definition of $\Sigma$}\\[1em]
&= (k^2+k) + 2(k+1)\tag{by $P(k)$, the ind. hyp.}\\[1em]
&= (k+1)^2+(k+1),\tag{manipulate expression}
\end{align}
we end up with the right-hand side of $P(k+1)$. 
Thus, by mathematical induction, the statement $P(n)$ holds for all $n\geq 1$. $\blacksquare$
A: I believe this might help you. 
Hint: $\sum2n=2\sum n$, but $\sum n=\frac{n^{2}+n}{2}$.
