1. ## Show that...

Show that $\frac{1}{r!}-\frac{1}{(r+1)!} = \frac{r}{(r+1)!}$

thanks!

2. Originally Posted by BabyMilo
Show that $\frac{1}{r!}-\frac{1}{(r+1)!} = \frac{r}{(r+1)!}$

thanks!
(r+1)! = (r+1) * r!

So lcm(r!, (r+1)!) divides (r+1)!

We can add the fractions on the LHS using the denominator (r+1)!

You should quickly see why it works out.

By the way, LHS means left hand side of the equation, and if you didn't understand the line about lcm, it's not terribly important, as long as you can see why (r+1)! can be used as a common denominator.

Edit:

Actually, the stronger statement

lcm(r!, (r+1)!) = (r+1)!

holds. I just wrote the weaker statement because I knew it was safe, without having to think. But obviously lcm(a, b) >= max(a, b) so it was a pretty silly precaution.

3. Originally Posted by undefined
(r+1)! = (r+1) * r!

So lcm(r!, (r+1)!) divides (r+1)!

We can add the fractions on the LHS using the denominator (r+1)!

You should quickly see why it works out.

By the way, LHS means left hand side of the equation, and if you didn't understand the line about lcm, it's not terribly important, as long as you can see why (r+1)! can be used as a common denominator.
thank you but how do i know (r+1)! = (r+1) * r! ?

4. Originally Posted by BabyMilo
thank you but how do i know (r+1)! = (r+1) * r! ?
Consider

5! = 5 * 4 * 3 * 2 * 1

6! = 6 * 5 * 4 * 3 * 2 * 1

It should be apparent that 6! = 6 * 5!

5. Originally Posted by BabyMilo
thank you but how do i know (r+1)! = (r+1) * r! ?
Because that follows directly from the definition of n!.

6. Originally Posted by HallsofIvy
Because that follows directly from the definition of n!.
Yes there are multiple definitions of n!, one of them is recursive. I didn't bring it up in case the OP was only familiar with the non-recursive definition.