# Thread: wanted to check to make sure I did this partial fraction right..

1. ## wanted to check to make sure I did this partial fraction right..

i have done partial fractions, but we are getting into series, and I need to use partial fractions to split it up, and just wondered if I did this one right, as its been a little while since ive done them. Thanks!

2. Originally Posted by BCHurricane89
i have done partial fractions, but we are getting into series, and I need to use partial fractions to split it up, and just wondered if I did this one right, as its been a little while since ive done them. Thanks!

nope, that's incorrect i'm afraid

$\displaystyle \frac {2n + 1}{n^2(n + 1)^2} = \frac An + \frac B{n^2} + \frac C{n + 1} + \frac D{(n + 1)^2}$

3. really? I thought the numerator had to be 1 power less than the denominator, hence why I have the bn+c and the en+f over the squared terms.

4. Originally Posted by BCHurricane89
really? I thought the numerator had to be 1 power less than the denominator, hence why I have the bn+c and the en+f over the squared terms.
that is (technically) true. in fact, that is how we started:

$\displaystyle \frac {2n + 1}{n^2(n + 1)^2} = \frac {An + B}{n^2} + \frac {C(n + 1) + D}{(n + 1)^2} = \frac An + \frac B{n^2} + \frac C{n + 1} + \frac D{(n + 1)^2}$

this is just the rule, here's the trick. look at the degree of the "inside" function. that is, the function that is raised to the power, that is the function your numerator must be one degree less than. consider the illustration below

$\displaystyle \frac {f(x)}{[g(x)]^n} = \frac {p_1(x)}{g(x)} + \frac {p_2(x)}{[g(x)]^2} + \cdots + \frac {p_n(x)}{[g(x)]^n}$

in the above, $\displaystyle f(x)$ is of lower degree than $\displaystyle g(x)$, and each $\displaystyle p_n(x)$ is a general polynomial function that is ONE degree less than $\displaystyle g(x)$.

so, for example, if it were $\displaystyle \frac {2n + 1}{n^2(n^2 + 1)^2}$, you would write

$\displaystyle \frac {2n + 1}{n^2(n^2 + 1)^2} = \frac An + \frac B{n^2} + \frac {Cn + D}{n^2 + 1} + \frac {En + F}{(n^2 + 1)^2}$

in the last two fractions, the "inside" function was a quadratic, so the numerators had to be linear.

if it were $\displaystyle \frac {2n + 1}{n^2(n^3 + 1)^2}$, then you would write

$\displaystyle \frac {2n + 1}{n^2(n^3 + 1)^2} = \frac An + \frac B{n^2} + \frac {Cn^2 + Dn + E}{n^3 + 1} + \frac {Fn^2 + Gn + H}{(n^3 + 1)^2}$

in the last two fractions, the "inside" function was a cubic, so the numerators had to be quadratic

got it?

5. ahh ok, I see how u did it...so the trick is to look at the power of the inside fraction

6. Originally Posted by BCHurricane89
ahh ok, I see how u did it...so the trick is to look at the degree of the inside function
the "inside" function of the denominator, yes. your numerator must be one degree less than that

7. it's worth to mention that $\displaystyle \frac{2n+1}{n^{2}(n+1)^{2}}=\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}.$

8. Originally Posted by Krizalid
it's worth to mention that $\displaystyle \frac{2n+1}{n^{2}(n+1)^{2}}=\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}.$
of course it is