# Can someone explain me this please?

## $\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$

Jun 6, 2017

See below.

#### Explanation:

Take the right hand side of the equation:

$\frac{1}{n + 1} - \frac{1}{n + 2}$

Create a common denominator so you can subtract the 2 fractions. They common denominator is $\left(n + 1\right) \left(n + 2\right)$

$\frac{n + 2}{n + 2} \left(\frac{1}{n + 1}\right) - \frac{n + 1}{n + 1} \left(\frac{1}{n + 2}\right)$

$\frac{\left(n + 2\right) - \left(n + 1\right)}{\left(n + 1\right) \left(n + 2\right)}$

Be careful when distributing the minus sign.

$\frac{n + 2 - n - 1}{\left(n + 1\right) \left(n + 2\right)}$

$\frac{1}{\left(n + 1\right) \left(n + 2\right)}$

Jun 6, 2017

Here's what I got.

#### Explanation:

I'm guessing that you want to know how to use partial-fraction decomposition to get

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$

The first thing to notice here is that the denominator of the original fraction can be factored as a product of linear factors, i.e. factors that have the highest power of the variable equal to $1$.

$\left(n + 1\right) = {n}^{1} + 1 \to$ linear factor

$\left(n + 2\right) = {n}^{1} + 2 \to$ linear factor

This means that you can write the original fraction as a sum of two fractions, one that has the first factor as a denominator and the other that has the second factor as a denominator.

$\frac{1}{\textcolor{b l u e}{\left(n + 1\right)} \textcolor{p u r p \le}{\left(n + 2\right)}} = \frac{A}{\textcolor{b l u e}{\left(n + 1\right)}} + \frac{B}{\textcolor{p u r p \le}{\left(n + 2\right)}}$

To find the values of $A$ and $B$, multiply the first fraction by 1 = (n+2)/(n+2) and the second fraction by $1 = \frac{n + 1}{n + 1}$, where $n \ne \left\{- 1 , - 2\right\}$.

This will get you

$\frac{A}{n + 1} \cdot \frac{n + 2}{n + 2} = \frac{A \cdot \left(n + 2\right)}{\left(n + 1\right) \left(n + 2\right)} = \frac{A n + 2 A}{\left(n + 1\right) \left(n + 2\right)}$

$\frac{B}{n + 2} \cdot \frac{n + 1}{n + 1} = \frac{B \cdot \left(n + 1\right)}{\left(n + 1\right) \left(n + 2\right)} = \frac{B n + B}{\left(n + 1\right) \left(n + 2\right)}$

You now have

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{A n + 2 A}{\left(n + 1\right) \left(n + 2\right)} + \frac{B n + B}{\left(n + 1\right) \left(n + 2\right)}$

All three fractions have the same denominator, which means that you can write

$1 = A \cdot n + 2 A + B \cdot n + B$

This is equivalent to

$1 = \left(A + B\right) \cdot n + \left(2 A + B\right)$

or

$\textcolor{red}{0} \cdot n + \textcolor{\mathrm{da} r k g r e e n}{1} = \textcolor{red}{\left(A + B\right)} \cdot n + \textcolor{\mathrm{da} r k g r e e n}{\left(2 A + B\right)}$

You can thus say that

$\left\{\begin{matrix}A + B = 0 \\ 2 A + B = 1\end{matrix}\right.$

Use the first equation to write

$A = - B$

Plug this into the second equation to get

$2 \cdot \left(- B\right) + B = 1$

$- B = 1 \implies B = - 1$

This means that

$A = - \left(- 1\right) = + 1$

Therefore, you can write the original fraction as

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} + \frac{- 1}{n + 2}$

which is equivalent to

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$

You can double-check your work by starting from the right-hand side of the equation

$\frac{1}{n + 1} - \frac{1}{n + 2}$

The common denominator will be

$\left(n + 1\right) \left(n + 2\right)$

which means that you have

$\frac{1}{n + 1} \cdot \frac{n + 2}{n + 2} - \frac{1}{n + 2} \cdot \frac{n + 1}{n + 1}$

$\frac{n + 2}{\left(n + 1\right) \left(n + 2\right)} - \frac{n + 1}{\left(n + 1\right) \left(n + 2\right)}$

At this point, you can subtract the two numerators to find

$n + 2 - \left(n + 1\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} + 2 - \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} - 1 = 1$

Therefore, you once again have

$\frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{\left(n + 1\right) \left(n + 2\right)}$