# Can someone explain me this please?

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#1/((n+1)(n+2)) = 1/(n+1) - 1/(n+2)#

##### 2 Answers

See below.

#### Explanation:

Take the right hand side of the equation:

Create a common denominator so you can subtract the 2 fractions. They common denominator is

Be careful when distributing the minus sign.

Here's what I got.

#### Explanation:

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

#1/((n+1)(n+2)) = 1/(n+1) - 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

#(n+1) = n^1 + 1 -># linear factor

#(n+2)= n^1 + 2 -># 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.

#1/(color(blue)((n+1))color(purple)((n+2))) = A/color(blue)((n+1)) + B/color(purple)((n+2))#

To find the values of

This will get you

#A/(n+1) * (n+2)/(n+2) = (A * (n+2))/((n+1)(n+2)) = (An + 2A)/((n+1)(n+2))#

#B/(n+2) * (n+1)/(n+1) = (B * (n+1))/((n+1)(n+2)) = (Bn + B)/((n+1)(n+2))#

You now have

#1/((n+1)(n+2)) = (An + 2A)/((n+1)(n+2)) + (Bn + B)/((n+1)(n+2))#

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

#1 = A * n + 2A + B * n + B#

This is equivalent to

#1 = (A + B) * n + (2A + B)#

or

#color(red)(0) * n + color(darkgreen)(1) = color(red)((A + B)) * n + color(darkgreen)((2A + B))#

You can thus say that

#{(A + B = 0), (2A + B = 1) :}#

Use the first equation to write

#A = -B#

Plug this into the second equation to get

#2 * (-B) + B = 1#

#-B = 1 implies B = -1#

This means that

#A = - (-1) = +1#

Therefore, you can write the original fraction as

#1/((n+1)(n+2)) = 1/(n+1) + (-1)/(n+2)#

which is equivalent to

#1/((n+1)(n+2)) = 1/(n+1) - 1/(n+2)#

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

#1/(n+1) - 1/(n+2)#

The common denominator will be

#(n+1)(n+2)#

which means that you have

#1/(n+1) * (n+2)/(n+2) - 1/(n+2) * (n+1)/(n+1)#

#(n+2)/((n+1)(n+2)) - (n+1)/((n+1)(n+2))#

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

#n+ 2 - (n+1) = color(red)(cancel(color(black)(n))) + 2 - color(red)(cancel(color(black)(n))) - 1 =1#

Therefore, you once again have

#1/(n+1) - 1/(n+2) = 1/((n+1)(n+2))#