How do I get from $\frac13n(n+1)(n+2)+(n+2)(n+1)$ to $\left(\frac13n + 1\right)(n+1)(n+2)$?

Question

In my coursebook, on of the steps during a mathematical induction is the following:

$$\frac13n(n+1)(n+2)+(n+2)(n+1) = \left(\frac13n+1\right)(n+1)(n+2).$$

I have tried to reconstruct the authors line of thought, but I end up at $$\frac13n(n+1)(n+2)+(n+2)(n+1) = \left(\frac13n^2+\frac13n\right)\left(\frac13n^2+\frac23n\right)+(n+2)(n+1)$$

What am I missing here?

Any help would be greatly appreciated!

Answer 1

Just factor out $(n+1)(n+2)$.

$$\frac13 n \color{red}{(n+1)(n+2)} + \color{red}{(n+1)(n+2)}=\left(\frac13 n+1\right)\color{red}{(n+1)(n+2)} $$

Also notice that in your working, you have involved $n^4$, which should not happen since we are only working with addition of polynomials of degree at most $3$.

Answer 2

Your mistake is here $\to$

$$\require{cancel}\frac13n(n+1)(n+2)+(n+2)(n+1) = \left(\frac13n^2+\frac13n\right)\underbrace{\cancel{\left(\frac13n^2+\frac23n\right)}}_{\color{red}{\text{should stay }(n+2)}}+(n+2)(n+1)$$

You have done the following $abc=(ab)(ac)$ instead of $abc=(ab)c$

Answer 3

if it helps, designate $A=n+1$ and $B=n+2$

Now it's $\frac{1}{3}nAB+AB$

Which is just $AB[\frac{1}{3}n + 1]$