Question

Can someone explain me this please?

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

Answer 1

See below.

Explanation:

Take the right hand side of the equation:

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

Create a common denominator so you can subtract the 2 fractions. They common denominator is `(n+1)(n+2)`

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

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

Be careful when distributing the minus sign.

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

`1/((n+1)(n+2))`

Answer 2

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 `1`.

`(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 `A` and `B`, multiply the first fraction by #1 = (n+2)/(n+2)# and the second fraction by `1 = (n+1)/(n+1)`, where `n!= {-1, -2}`.

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))`