partial fractions

**guyonfire89** · September 20th 2010, 05:16 PM

I'm a little rusty with my partial fractions so I just need a refresher.

The equation is $1/(x+1)(x+2)(x+4)$ I have
$A/x+1 + B/x+2 + C/x+4$
Then common denominators

$(x+1)(x+2)(x+4)= A(x+2)(x+4) + B(x+1)(x+4) + C(x+1)(x+2)$

After this can I do where, x=-2 for example? Cause then I will get 0=-2B
And then keep plugging in values for x? Or is there another way which I complete forgot about

**Prove It** · September 20th 2010, 05:43 PM

Yes there is another way.

For starters, you should actually have gotten

$\frac{A(x + 2)(x + 4) + B(x + 1)(x + 4) + C(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 4)} = \frac{1}{(x + 1)(x + 2)(x + 4)}$

$A(x + 2)(x + 4) + B(x + 1)(x + 4) + C(x + 1)(x + 2)=1$

$A(x^2 + 6x + 8) + B(x^2 + 5x + 4) + C(x^2 + 3x + 2) = 1$

$Ax^2 + 6Ax + 8A + Bx^2 + 5Bx + 4B + Cx^2 + 3Cx + 2C = 1$

$(A + B + C)x^2 + (6A + 5B + 3C)x + 8A + 4B + 2C = 0x^2 + 0x + 1$ .

This means

$A + B + C = 0$
$6A + 5B + 3C = 0$
$8A + 4B + 2C = 1$ .

Solve these equations simultaneously to find $A, B, C$ .

**guyonfire89** · September 20th 2010, 05:48 PM

thanks. Forgot about solving with coefficients

**guyonfire89** · September 20th 2010, 06:10 PM

so I have the series from 1 to infinity of $1/3(x+1) -1/5(x+2) +1/6(x+4)$

How do I solve the sum from there? I tried telescoping, and it doesn't work

**TheCoffeeMachine** · September 20th 2010, 06:20 PM

When you get to $1 \equiv a(x+2)(x+4)+b(x+1)(x+4)+c(x+1)(x+2)$ , you can put $x = -1, -2, -4$ to respectively find $a, b, c$ (instead of solving the system from the coefficient relations).

**guyonfire89** · September 20th 2010, 06:32 PM

Originally Posted by TheCoffeeMachine

Ok. When you get to $1 \equiv a(x+2)(x+4)+b(x+1)(x+4)+c(x+1)(x+2)$ , you can put $x = -1, -2, -4$ to respectively find $a, b, c$ (instead of solving the system from the coefficient relations).

Will telescoping series work from there to find the sum?

**mr fantastic** · September 20th 2010, 07:19 PM

Originally Posted by guyonfire89

Will telescoping series work from there to find the sum?

IF the series is meant to be the sum of what you posted in post #1 (which I corrected because of the typos) , then yes, it is a telescoping series. To realise this you need to:

1. Find the correct partial fraction decomposition.

2. Write out the first few terms ( I suggest at least the first four terms) and start cancelling in the usual way.

If you take greater care with the details you will probably have better luck in solving questions like this one.

**Soroban** · September 20th 2010, 08:56 PM

Hello, guyonfire89!

An error in your thiird statement.

$\text{The fraction is: }\;\dfrac{1}{(x+1)(x+2)(x+4)}$

$\text{I have: }\;\dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{x+4}$

We have: . $\displaystyle \frac{1}{(x+1)(x+2)(x+4)} \;=\;\frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+4}$

Multiply through by the LCD:

. . $\displaystyle 1 \;=\;A(x+2)(x+4) + B(x+1)(x+4) + C(x+1)(x+2)$
m↑

$\begin{array}{cccccccccccc}<br /> \text{Let }x = \text{-}1\!: & 1 &=& A(1)(3) + B(0) + C(0) & \Rightarrow & A &=& \frac{1}{3} \\ \\[-3mm]<br /> \text{Let }x = \text{-}2\!: & 1 &=& A(0) + B(\text{-}1)(2) + C(0) & \Rightarrow & B &=& \text{-}\frac{1}{2} \\ \\[-3mm]<br /> \text{Let }x = \text{-}4\!: & 1 &=& A(0) + B(0) + C(\text{-}3)(\text{-}2) & \Rightarrow & C &=& \frac{1}{6} \end{array}$

Therefore: . $\displaystyle \frac{1}{(x+1)(x+2)(x+4)} \;=\;\frac{\frac{1}{3}}{x+1} + \frac{\text{-}\frac{1}{2}}{x+2} + \frac{\frac{1}{6}}{x+4}$

. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle =\;\frac{1}{6}\left(\frac{2}{x+1} - \frac{3}{x+2} + \frac{1}{x+4}\right)$

**guyonfire89** · September 20th 2010, 09:37 PM

thanks guys, but when I try to prove its a telescoping series, only the first two terms cancel out. I've tried out the first 4 terms

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