Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.
Inductive hypothesis:
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ is true when $n = k$ and $k > 1$
Therefore: $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1}$$
Inductive step:
Prove that $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+1+1} = \frac{k+1}{k+2}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+2}$$
What I am confused about is where the $\frac{1}{(k+1)(k+2)}$ comes from in the first line of the inductive step. Can someone please explain this in a little more detail? The source of the answer explains it as "break last term from sum", but I am unclear on what that means.

The base case $n=1$ is true.  Assume that for some $n> 1$ 
$$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$$
Then, we have
$$\begin{align}
\sum_{k=1}^{n+1}\frac{1}{k(k+1)}&=\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n(n+2)+1}{(n+1)(n+2)}\\\\
&=\frac{(n+1)^2}{(n+1)(n+2)}\\\\
&=\frac{n+1}{n+2}
\end{align}$$
which completes the proof by induction.
