Prove by induction on $n$ that $(1)(2)+(2)(3)+...+n(n+1)={1\over 3}n(n+1)(n+2)$

Question

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?

Answer 1

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

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

Answer 2

They are equal .

Indeed, we have $\frac{1}{3} (k+1)(k+2) $ common in both the terms. Take the common term out, then, we have : $ \frac{1}{3} (k+1)(k+2) \cdot [k + 3]$, where $ k$ comes from the first term and $3$ from the second one. This is what the last line is.