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 $n\geq 1$. $\blacksquare$
