How does this seemingly-trivial simplification work?

Question

In a section on inductive proofs in the book Modelling Computing Systems: Mathematics for Computer Science (Muller, Struth) there is a simplification that is assumed to be trivial, but that I can't figure out.

It occurs in this step:

$$ \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \stackrel{?}= \frac{(k+1)(k+2)(k+3)}{3} $$

How does one get from the first expression to the second?

Answer 1

$$ \begin{align} \frac{k(k+1)(k+2)}{3} + (k+1) (k+2) &= \frac{\color{red}{k}(k+1)(k+2) + \color{red}{3}(k+1)(k+2)}{3}\\ &= \frac{(k+1)(k+2)(\color{red}{k}+\color{red}{3})}{3} \end{align} $$

Answer 2

Hint : Multiply $(k+1)(k+2)$ with $3$ and add $k(k+1)(k+2)$

Then factor out $(k+1)(k+2)$

Answer 3

Hint: $$(k+1)(k+2)=\frac{3\cdot(k+1)(k+2)}{3}.$$ Now combine the fractions and factor out $(k+1)(k+2)$.

Answer 4

Rewrite the second term with a denominator of 3.

$$(k+1)(k+2)=\frac{3(k+1)(k+2)}{3}$$

Add the two fractions and factor out the common $(k+1)(k+2)$.

$$\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}=\frac{(k+1)(k+2)(k+3)}{3}$$

There you go!

Answer 5

If boils down to

$$\begin{align}\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) &=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\\ &=\frac{(k+1)(k+2)(k+3)}{3} \end{align}$$

Where the last step is just factoring$(k+1)(k+2)$ in the numerator.