<s>[INST] proof by induction (summations) I'm trying to use induction to prove that 
$$\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+...+\frac{1}{(n)(n+1)}=1-\frac{1}{n+1}.$$
This is what I have so far:
Base Case: $n=1$. We get $$\sum_{i=1}^{k} \frac{1}{(i)(i+1)} = 1-\frac{1}{(k+1)} \implies \frac{1}{2} = \frac{1}{2}.$$
End Goal (what we want to show): $$\sum_{i=1}^{k+1} \frac{1}{(i)(i+1)} = 1-\frac{1}{k+2}.$$
So, 
$$\begin{align}
&\sum_{i=1}^{k} \frac{1}{(i)(i+1)} + \frac{1}{(k+1)(k+2)}\\
&=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}\\
&= 1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
&= 1-\frac{(k+2)+1}{(k+1)(k+2)}
\end{align}$$
But I don't know where to go from here. I can't cancel the $k+2$'s because I need one in the denominator so the LHS = RHS, but if I expand the numerator to $(k+1)+2$ and then cancel the $k+1$'s I'm left with a $2$ in the numerator when I need a $1$. Any suggestions?
 [/INST] You are almost there. The trick goes as by writing 
$$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$$ so that 
$$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}=1-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}
$$
</s>