Can someone explain me this please?
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))