1. Summation: Method of differences.

Heres the question:

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

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
$\displaystyle \sum \frac{3r + 4}{ r(r+1)(r+2)}$to n. Where it starts r = 1

Okey, so, method of differences.

I start by putting in r = 1

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

Then r = 2

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

Then r = n-1

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

Then r = n

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

Then r = r + 1

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

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

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

Then i simplify and i get

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

And this is incorrect for the answer is

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

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.

2. Originally Posted by AshleyT
Heres the question:

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

I did this part since its simple algebra. .
How did you do this? It is wrong!

The correct identity is

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

Then the second part i'm stuck.

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

Okey, so, method of differences.\
$\displaystyle \sum_{r=1}^{r=n} \frac{3r + 4}{ r(r+1)(r+2)}$

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

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

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

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

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

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

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.

3. Originally Posted by Isomorphism
How did you do this? It is wrong!

The correct identity is

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

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

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

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

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

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

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

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

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

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

to

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

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

Tanks

4. Originally Posted by AshleyT
Oops, i meant 2/r sorry. Typed it out wrong!

How did you go from

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

to

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

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

Tanks

I just wrote $\displaystyle -\frac{1}{r+1}$ as $\displaystyle -\frac2{r+1} + \frac1{r+1}$ and then grouped $\displaystyle -\frac2{r+1}$ with $\displaystyle \frac2{r}$ and $\displaystyle \frac1{r+1}$ with $\displaystyle \frac1{r+2}$

5. Originally Posted by Isomorphism
I just wrote $\displaystyle -\frac{1}{r+1}$ as $\displaystyle -\frac2{r+1} + \frac1{r+1}$ and then grouped $\displaystyle -\frac2{r+1}$ with $\displaystyle \frac2{r}$ and $\displaystyle \frac1{r+1}$ with $\displaystyle \frac1{r+2}$

O i see.

Thankyou very much