Prove that every even perfect number is a triangular number. I know a triangular number is given by the formula $\frac{n(n+1)}{2}$
I also know that an even perfect number is given by $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.
Please help me to prove this.
 A: You should use different variables in the two expressions.  An even perfect number is then $2^{k-1}(2^k-1)$  You need to find an $n$ such that $\frac 12n(n+1)=2^{k-1}(2^k-1)$.  I find the right side quite suggestive. Therefore $n=2^{k}-1$
A: You could think as follows:
Every triangular number you can represent as follows:
$$
\bullet\\
\ \ \ \ \bullet\ \bullet\\
\ \ \ \ \ \ \ \ \ \bullet\ \bullet\ \bullet \\
\ \ \ ...
$$
that is $1+2+3+\ldots +n=\frac{n(n+1)}{2}$.
Multiplying this by $2$ gives a rectangle
$$
\bullet\ \circ \ \circ \   \circ\\
\bullet\ \bullet \ \circ \ \circ\\
\bullet\ \bullet\ \bullet \ \circ\\
$$
with one side exactly one unit longer than the other. 
So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.
Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed 
$$
\frac{n(n+1)}{2}=2^{k-1}(2^k-1),
$$
so 
$$
\text{$a$ is perfect}\Rightarrow \text{$a$ is triangular}.
$$
