# Algebraic Manipulation

What is the best method to get the LHS equal to RHS?

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

Thank you.

• There's just one equation here. If you want to see why the left side is equal to the right, factor out $(n+1)(n+2)$. Apr 20, 2015 at 17:00
• Thanks that's what I meant to say. Apr 20, 2015 at 17:02

$$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3} \tag{\star}$$

Notice that the product $(n+1)(n+2)$ is common to all the terms in $(\star)$. If $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$, then $(n+1)(n+2) \neq 0$ and you can $\color{blue}{\text{simplify}}$ by $(n+1)(n+2)$ as follows :

$$\frac{\require{cancel} n\color{blue}{\cancel{(n+1)(n+2)}}}{3} + \require{cancel} \color{blue}{\cancel{(n+1)(n+2)}} = \frac{\require{cancel} \color{blue}{\cancel{(n+1)(n+2)}}(n+3)}{3} \tag{\star}$$

It gives :

$$\frac{n}{3} + 1 = \frac{n+3}{3} = \frac{n}{3} + 1.$$

which is true for all $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$.

Since $1=\frac 33$, one has \begin{align}\frac{n(n+1)(n+2)}{3}+\color{blue}{1}\cdot (n+1)(n+2)&=\frac{n(n+1)(n+2)}{3}+\frac{\color{blue}{3}(n+1)(n+2)}{\color{blue}3}\\&=\frac{n\color{red}{(n+1)(n+2)}}{\color{red}{3}}+\frac{3\color{red}{(n+1)(n+2)}}{\color{red}{3}}\\&=\color{red}{\frac{(n+1)(n+2)}{3}}(n+3)\end{align}

Firstly, take $(n+1)(n+2)$ out as a factor:

$((n+1)(n+2))[\frac n3 +1] = \frac{n+3}{3} (n+1)(n+2)$

Then divide through by $(n+1)(n+2)$:

$(\frac n3 +1) = \frac{n+3}{3}$

Multiply both sides by $3$:

$n+3 = n+3$

Therefore LHS = RHS