# Method of Differences

1. Nov 17, 2008

### thomas49th

1. The problem statement, all variables and given/known data
I'm given the identity:
$$\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}$$

EDIT: from r=1 to n on the sum sign
2. Relevant equations

I know I'm ment to use the method of differences and I know that

$$\sum(r) = \frac{n(n+1)}{2}$$

3. The attempt at a solution

I presume you need to find out how you can relate the 3 fractions to the summation of r?

Thanks :)

Last edited: Nov 17, 2008
2. Nov 17, 2008

### tiny-tim

Hi thomas49th!

Hint:

i] r3 - r = … ?

ii] with a view to using differences, 1/r(r+1) = … - … ?

3. Nov 17, 2008

### thomas49th

sorry to of wasted your time but the orginal question was

$$\sum{\frac{r^{3}-r+1}{r(r+1)} \equiv r-1 + \frac{1}{r} - \frac{1}{r+1}$$

there is no 1/2

sorry!

Errr can I combine it to this:
$$\sum (r) - 1 + \sum(\frac{1}{r} - \frac{1}{r+1})$$

does that help at all?

$$\sum(\frac{1}{r} - \frac{1}{r+1})$$
simplifies to $$\sum(\frac{1}{r^{2} + r})$$

Thanks! :)

4. Nov 17, 2008

### tiny-tim

So ∑ 1/(r2 + r) = … ?
erm … I could see it was wrong! … in fact, I still think it's wrong!!
Not following you.

The method of difference involves subtracting the answer for r = n from the answer for r = n-1.

5. Nov 17, 2008

### thomas49th

It's question number 4

I could do part a) that's easy just part part b) is giving me trouble.

At first I tried $$\sum (r) - 1 + \frac{1}{\sum (r)} - \frac{1}{\sum (r) + 1}$$

but that's wrong?

Thanks :)

6. Nov 17, 2008

### tiny-tim

The ∑n part is right.

The 1 should be ∑1.

The rest is just ∑(1/r - 1/r+1) …

that's already in differences, isn't it? …

so its sum is … ?

7. Nov 17, 2008

### thomas49th

is the sum:
$$\frac{n(n+1)}{2} - n + \frac{2}{n(n+1)} - \frac{2}{n(n+1) + 2n}$$

is that right?

is the sum of 1 = n?

8. Nov 17, 2008

### tiny-tim

The first two terms are right.

As I said before: The rest is just ∑(1/r - 1/r+1)

Look … that's (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + …

can you see what's happening?

this is why it's called the method of differences, and why it works!

9. Nov 17, 2008

### thomas49th

I know the trick when things cancel out.
So are you substituting values 1 to n into 1/2 - 1/r+1?

how was I ment to spot that you substitute values into there and not the first part?

Thnanks :)

10. Nov 17, 2008

### tiny-tim

Exactly!
Because the first part was easy … it was just ∑(r + 1) …

but the second part was ∑ 1/r(r+1), which obviously needs a trick!

11. Nov 18, 2008

### thomas49th

Ahh yes I got it

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

Thank you ever so much :)