Prove by induction on $n$ that $(1)(2)+(2)(3)+...+n(n+1)={1\over 3}n(n+1)(n+2)$ So after testing myself with this question, I was unable to solve it. I was able to prove the base case $n=1$, but I was pretty lost on the induction step. I took a look at the solution and here it is: Solution to problem
I understand it up until the point in the $P(k+1)$ step where it says:
$={1\over 3}(k+1)(k+2)(k)+{1\over 3}(k+1)(k+2)(3)$
$={1\over 3}(k+1)(k+2)[k+3]$
I don't see how they have made this jump. It certainly seems like a bigger jump than any of the other lines. What is the process here? I'm not great at factorising... but if I were to replace $k$ with a number, I'm pretty sure these lines would not be equal! So what am I missing?
Do the square brackets carry some kind of special notation in this situation?

This is related to the distributive property which basically states that $\rm \color{red}ab+\color{red}ac=\color{red}a\cdot(b+c)$.
In your example 

$$\rm \color{blue}{\frac13(k+1)(k+2)}\cdot k+ \color{blue}{\frac13(k+1)(k+2)}\cdot3=\color{blue}{\frac13(k+1)(k+2)}\cdot(k+3)$$

