Find the partial fraction decomposition for 1/x(x+1) and use the ...

Question

Find the partial fraction decomposition for 1/x(x+1) and use the result to find the following sum:

$Mathematical series showing partial fractions: 1/1·2 + 1/2·3 + ... + 1/99·100.$

Accepted Answer

Hello. Today, we're going to be performing the partial fraction decomposition on the given expression. And then using our result to evaluate the following addition. So let's go ahead and start by doing the partial fraction decomposition on our given expression. And the expression given to us is three over Y times Y plus three. Now, before we start our partial fraction decomposition, we always want to make sure that our denominator is fully factored in this case here. There's no way to simplify why by itself and there's no way to simplify Y plus three. So our current denominator is fully factored. Now, let's just go ahead and take a look at each factor in our given denominator, why is a linear quantity as it is a variable with the exponent of one. So why is going to produce the partial fraction of A over Y and Y plus three is a linear quantity as the leading coefficient has an exponent of one. So that's going to produce the partial fraction of B Over Y-plus three. Now, what we want to go ahead and do is solve our missing coefficients A and B. But in order to do that, we need to first get rid of all of the denominators are given in their given equation. And in order to do that we can multiply everything by the lowest common denominator. And in this case here, the L C D is going to be Y times Y plus three. So let's just go ahead and write that out. We have our original expression times the L C D, which is why times Y plus three over one is equal to the first partial fraction which is A over Y times the L C D. Which is why times Y plus 3/1 plus our second partial fraction which is B over Y plus three times Y times Y plus 3/1. Multiplying the left hand side of this equation by the L. C. D. Is going to completely cancel out the denominator. Multiplying our first partial fraction by the L C. D is going to cancel the denominator. But leave us with Y plus three in the numerator and multiplying the L. C. D. To our second partial fraction is going to cancel the denominator but leave us with Y in the numerator. So what we have is three is equal to A times Y plus three plus B, times Y, which is B. Y. Now we want to go ahead and simplify the right hand side. And the only thing we have to do is just distribute a into the quantity Y plus three. And doing this is going to give us eight times Y plus three A plus B. Y. And what we're going to do is group up all of our white terms together and set aside the constants and doing this is going to give us three is equal to eight times Y plus B, Y plus three A. And we can go ahead and factor out A. Y. From this quantity here and that's going to give us three is equal to y times A plus B plus three A. Now we can go ahead and use this equation to essentially create a system of equations that allow us to solve for A. And B. And in order to do that we're going to set the coefficients of all of our white terms equal to each other and all the constants equal to each other. However, if you notice we don't have a wide term on the left hand side of this equation, but we can go ahead and add one as long as it has a coefficient of zero, so we have zero, Y plus three is equal to everything else. So our first equation is going to be the coefficients of Y on the right hand side which is a plus B equal to the coefficient of Y on the left hand side which is zero. So we have a plus B is equal to zero for our first equation. And for a second equation we're just gonna go ahead and set the constant equal to each other. So our second equation is going to give us three A is equal to three. Now this is pretty easy to solve because for our second equation we just have to divide both sides by three and what that gives us is A is equal to one and we can go ahead and take this value of A. And substitute into our first equation to give us one plus B is equal to zero and subtracting one to both sides of this equation is going to give us B is equal to negative one and now we can just go ahead and put that into our original partial fractions that we came up with earlier and doing this is going to give us a over Y or in this case one over Y plus B over Y plus three, which is going to be negative one over Y plus three. But we can move the negative to the front of the fraction. So we have minus one over Y plus three. And this is the complete partial fraction decomposition for our given expression. Now we want to go ahead and use this value to evaluate this given summation. And in order to evaluate this, let's just take a look at what our original expression was. Our original expression was three over Y times Y plus three. Let's go ahead and just plug in some values for why? If we plug in, Y is equal to one, we get 3/1 times one plus three, Which is one times 4. If we plug in Y is equal to two, we get 3/2 times two plus three, which is five. And if we plug in, why is equal to three we get 3/3 times three plus three which is six. Now if you notice these values of why that we're getting exactly match the given summation up above. So this original expression Is a proper representation of the given summation. However, the first term is when Y is equal to one and the last term is when Y is equal to 99. So we can go ahead and represent this as the summation. So we write our sigma notation with, the first term being y is equal to one and the last term being of three over Y times Y plus three. But we have a substitution for this stuff. We came up with it by performing partial fraction decomposition and we know that three over y times Y plus three is equal to one over y minus one over y plus three. So we can rewrite this summation As the summation of 99, why? It is equal to one of one over y minus one over y plus three. And evaluating the summation by plugging it into the calculator is approximately going to give us the value of 1. to of course this being estimated to five decimal places and this is going to be the solution to this given problem and that means that the answer to this problem is going to be c. So I hope this video helps you in understanding how to solve this type of problem and I'll go ahead and see you all in the next video

Find the partial fraction decomposition for 1/x(x+1) and use the result to find the following sum: $Mathematical series showing partial fractions: 1/1·2 + 1/2·3 + ... + 1/99·100.$

Key Concepts

Partial Fraction Decomposition

Summation of Series

Telescoping Series

Watch next