Summation: Method of differences.

**AshleyT** · May 20th 2008, 02:53 AM

Heres the question:

Show that
$\frac{2}{r-1} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}$

I did this part since its simple algebra.

.

Then the second part i'm stuck.

ii.) Hence write an expression in terms of n for
$\sum \frac{3r + 4}{ r(r+1)(r+2)}$ to n. Where it starts r = 1

Okey, so, method of differences.

I start by putting in r = 1

$\frac{2}{1} - \frac{1}{ 2} - \frac{1}{3}$

Then r = 2

$\frac{2}{ 2} - \frac{1}{3} - \frac{1}{4}$

Then r = n-1

$\frac{2}{n-1} - \frac{1}{n} - \frac{1}{n+1}$

Then r = n

$\frac{2}{n} - \frac{1}{n+1} - \frac{1}{n+2}$

Then r = r + 1

$\frac{2}{n+1} - \frac{1}{n+2} - \frac{1}{n+3}$

So then you add them all. Some cancel and im left with

$\frac{2}{1} + \frac{2}{2} + \frac{2}{n-1} + \frac{2}{n} + \frac{2}{n+1}$

Then i simplify and i get

$2 + 1 + \frac{2n(3n - 1)}{ n(n+1)(n-1)}$

And this is incorrect for the answer is

$2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2}$

Thankyou for any help you can offer.

How do i do all these maths things to make it easier to read?

EDIT: Nvm i found the tutorial . I'll edit this with it now.

**Isomorphism** · May 20th 2008, 03:18 AM

Originally Posted by AshleyT

Heres the question:

Show that
$\frac{2}{r-1} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}$

I did this part since its simple algebra.

.

How did you do this? It is wrong!

The correct identity is

$\frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}$

Then the second part i'm stuck.

ii.) Hence write an expression in terms of n for
$\sum \frac{3r + 4}{ r(r+1)(r+2)}$ to n. Where it starts r = 1

Okey, so, method of differences.\

$\sum_{r=1}^{r=n} \frac{3r + 4}{ r(r+1)(r+2)}$

$= \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=1}^{r=n} \left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=2}^{r= n+1} \left(\frac{1}{r} - \frac{1}{r + 1}\right)$

$= 2 - \frac2{n+1} + \frac12 - \frac1{n+2}$

$= 2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2} $

EDIT: Oh yay , I got the right answer

And so on.... I might have made arithmetic mistakes... but I hope you get the general idea.

**AshleyT** · May 20th 2008, 03:36 AM

Originally Posted by Isomorphism

How did you do this? It is wrong!

The correct identity is

$\frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} = \frac{3r + 4}{ r(r+1)(r+2)}$

$\sum_{r=1}^{r=n} \frac{3r + 4}{ r(r+1)(r+2)}$

$= \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=1}^{r=n} \left(\frac{1}{r+1} - \frac{1}{r + 2}\right)$

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) + \sum_{r=2}^{r= n+1} \left(\frac{1}{r} - \frac{1}{r + 1}\right)$

$= 2 - \frac2{n+1} + \frac12 - \frac1{n+2}$

$= 2 + 1 - \frac{1}{2} - \frac{2}{n+1} - \frac{1}{n+2} $

EDIT: Oh yay , I got the right answer

And so on.... I might have made arithmetic mistakes... but I hope you get the general idea.

Oops, i meant 2/r sorry. Typed it out wrong!

How did you go from

$\sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2} $

to

$ \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right) $

because, i can see the $2\frac{1}{r}$ but not how - $\frac{1}{r+ 1}$ goes in there?

Tanks

**Isomorphism** · May 20th 2008, 03:42 AM

Originally Posted by AshleyT

Oops, i meant 2/r sorry. Typed it out wrong!

How did you go from

$= \sum_{r=1}^{r=n} \frac{2}{r} - \frac{1}{r+ 1} - \frac{1}{r + 2}

to

$= \sum_{r=1}^{r=n} 2\left(\frac{1}{r} - \frac{1}{r+ 1}\right) +\left(\frac{1}{r+1} - \frac{1}{r + 2}\right)

because, i can see the $2\left(\frac{1}{r}$ but not how - $\frac{1}{r+ 1}\right)$ goes in there?

Tanks