Prove the following using induction $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
I'm new to induction, but this is what I cam up with so far.
$$1 - \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{k+2+k}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2(k+1)}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2}{k(k+2)} = 1 - \frac{1}{k+2}$$
Where did I go wrong, or where do I go from here?
EDIT:
After some advice from Patrick (you're awesome!) I have...
$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
    $$1 - \frac{k + 2 + 1}{(k+1)(k+2)}$$
    $$1 - \frac{k + 3}{(k+1)(k+2)}$$
    $$\frac{(k+1)(k+2) - (k+3)}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k + k + 2 - k - 3)}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k - 1}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k - 1}{k^2 + 3k + 2}$$
Did I goof up somewhere again?
Thanks.

You went wrong in the first line. You want to prove that $$1-\color{red}{\frac{1}{k+1}} + \frac{1}{(k+1)(k+2)} = 1-\frac{1}{k+2}$$
Indeed, you're trying to show that $$\underbrace{\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{k(k+1)}}_{\text{$1-\frac{1}{k+1}$ by induction hypothesis}} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
Your second line then has a sign error:
$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{k + 2 \color{red}{-} 1}{(k+1)(k+2)}$$
