Prove $\frac{1}{n} =\frac{1}{n+1}+\frac{1}{n(n+1)}$ for all integers $n\in\Bbb Z$

Question

I'm pretty sure that we need induction, since it's the format I had to use for previous problems similar to this (it isn't specified that it HAS to be an inductive proof, either, if there is another way I can solve this one.) The base case checks out except for $-1$ and $0$, which leads to $1/0$ at some point in the formula, so I'm excluding those two from $n$. So I assume that $n=k$ holds as well and have to prove for $n=k+1$. I understand the format of an inductive proof.

I'm just not seeing how to manipulate the problem correctly for the induction to work. I've tried multiplying by $\frac{1}{k}$ but that messes up the last part of the equation, leading to $k^3$. Doing the $\gcd$ just makes the numerators messy in a way I can't easily clean up. I'm not sure what algebra I should be using to get the induction I need. What's the format for the solution to this problem?

Answer 1

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

Answer 2

$\frac{1}{n(n+1)}+\frac{1}{n+1}=(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{n+1}$.

Answer 3

Cut a square into $(n+1)$ pieces, with one piece across the full width (in green), then $n$ pieces in vertical slices - shown below is the scheme for 7 pieces.

Then you can see that the light-blue $\frac{1}{n+1}$ slice added to a $\frac 1n$ portion of the green $\frac{1}{n+1}$ slice - that is, $\frac{1}{n+1} + \frac 1n\frac{1}{n+1}$ - is a $\frac 1n$ slice of the square.