Proof by induction that $2 + 4 + 6 + cdots + 2n = n(n+1)$

Proving by induction. We'd like to show that

$2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.

A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:
Show that $S(1)$ is true.
Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.

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This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.

Answer

First, that answer is correct. If you note that $2+4+6+...+2n = 2*(1+2+3+...+n)=2 \frac{n*(n+1)}2 = n*(n+1)$

Second. Suppose that it is true, you have $2+4+6+...+2n = n*(n+1)$ This is $S(n)$
add 2(n+1) to both sides, giving
$$2+4+...+2n+2(n+1) = n*(n+1) + 2(n+1) = (n+2)(n+1)$$ which is $S(n+1)$