Can someone explain me this please?

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

2 Answers
JS
Jun 6, 2017

See below.

Explanation:

Take the right hand side of the equation:

1/(n+1)-1/(n+2)1n+1−1n+2

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

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

((n+2)-(n+1))/((n+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))n+2−n−1(n+1)(n+2)

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

Stefan V.
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

1/((n+1)(n+2)) = 1/(n+1) - 1/(n+2)1(n+1)(n+2)=1n+1−1n+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 11.

(n+1) = n^1 + 1 ->(n+1)=n1+1→ linear factor

(n+2)= n^1 + 2 ->(n+2)=n1+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))1(n+1)(n+2)=A(n+1)+B(n+2)

To find the values of AA and BB, multiply the first fraction by 1 = (n+2)/(n+2)1=n+2n+2 and the second fraction by 1 = (n+1)/(n+1)1=n+1n+1, where n!= {-1, -2}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))An+1⋅n+2n+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))Bn+2⋅n+1n+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))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 + B1=A⋅n+2A+B⋅n+B

This is equivalent to

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

or

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

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