Mathemathic induction proof

Question

I need to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$

Here is what I tried:

\begin{align} & \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}+\frac{1}{n+1(n+2)} \\[10pt] = {} & \frac{n}{n+1}+\frac{1}{n+1(n+2)} = \frac{(n+1)(n+2)+(n+1)}{n^2(n+2)} \end{align}

I am stuck from here, I feel there must be someway to simplify the last part to get to

$$\frac{n+1}{n+2}$$

Answer 1

$$\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}$$

Answer 2

Hint: write

$$\frac1{k(k+1)}=\frac1k-\frac1{k+1}$$

and now do the telescopic sum carefully.