TITLE: Math induction sum of even numbers
QUESTION [5 upvotes]: I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1)+2$
I don't know how to move forward from this.
Thanks.

REPLY [7 votes]: $2+4+6+........+2n = n(n+1)$
$2+4+6+........+2n+2(n+1) = n(n+1)+2(n+1)=(n+1)(n+2)$

REPLY [5 votes]: Suppose that
$$
2+4+6+\cdots+2n=n(n+1)
$$
add $2n+2$ to both sides:
$$
2+4+6+\cdots+2n+(2n+2)=n(n+1)+(2n+2)
$$
To finish the induction, you want the right side to be $(n+1)(n+2)$. Is it?
$$
n(n+1)+(2n+2)=n^2+n+2n+2=n^2+3n+2=(n+1)(n+2)
$$

REPLY [3 votes]: The statement is: $2+4+...+2n=n(n+1)$ for all $n$
Now, $n^2+3n+2=(n+1)(n+2)=(n+1)((n+1)+1)$ and that is what you want to have as a the result for $n+1$.