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.

"What I am confused about is where the 1/(k+1)*(k+2) comes from in the first line of the inductive step."
It comes because you are trying to evaluate for $n = k+1$
You want to prove the statement  $\frac 1{1*2} + .....+ \frac 1{(n-1)(n)} + \frac 1{n*(n+1)} = \frac n{n+1}$.
For $n = k$  if you replace $n$ with $k$ you are assuming $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$.
If you replace $n$ with $k + 1$ you get the statement:
$\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac {k+1}{k+2}$.
And that is the statement you wish to prove.  
We are assuming  $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$
So $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac k{k+1} + \frac 1{(k+1)(k+1)}$
You just need to prove that $\frac k{k+1} + \frac 1{(k+1)(k+1)}= \frac {k+1}{k+2}$.
.....
Anyway, you certainly did not transcribe to proof correctly.
