XP Math Games Arcade

**Sillysidley** · 07-31-2007

Quote:

Originally Posted by Archive

$\frac{1}{2}$

Time to test the multinomial theorem...

What is $(x+y+z+a+b)^4$ ?

Let's try reviving this again.
It'd be a waste of time for me to actually do go through each exponent one by one, but here is the multinomial theorem:http://en.wikipedia.org/wiki/Multinomial_theorem
Try posting one with only x,y,z, much faster for peole to do.
REMEMBER, MIDDLE SCHOOL (or State Mathcount) LEVEL ONLY (seriously let's just do state mathcount level, good practice)

Find the greatest number that the sum of 3 consecutive odd numbers must be divisible by

08-01-2007

It's 3, obviously.

Find the greatest number than cannot be expressed by sums of 2, 3, 4, 5, 6, 7, 8, and 9.

**Sillysidley** · 08-01-2007

Well, 2-9 can all be made (obviously) and you can get any any number higher by adding multiples of 10 (by adding 5s)
so 1.

Find $\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+...+<br /> \frac {1}{499*500}$

Oh, and if anyone cares, I meant to ask whats the largest number that must be a factor of any four consecutive odd numbers. (Three is too simple.)

08-01-2007

Um, what's the "c" doing there? Are you sure you typed the LaTeX correctly?

**Sillysidley** · 08-01-2007

Looks like I had to go to the next line for some reason..
Oh, I forgot to metion, this isone of my favorite mathcount problems. It has two different approaches (One with a formula, one with a kind of series.)

**Sillysidley** · 08-04-2007

I'm gonna post the solution by 6:00 if no one answers by that time...

**Sillysidley** · 08-04-2007

Eh, close enough
The first one uses a telescoping series (basically where a bunch of things cancel out) and the second has a formula (and just to get better at proofs, I wanna prove it)

Show more:

We want all the terms except the first and last to cancel. So, we express it as:
$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{499}-\frac{1}{500}=1-\frac{1}{500}=\frac{499}{500}$

2nd:

Show more:

There's a formula that says the sum of n numbers of this kind sum up to $\frac{n}{n+1}$ . In other words, $\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n*(n+1)}= \frac{n}{n+1}$ We can then easily see the answer is $\frac{499}{500}$ .

Proof: Easy enough to prove by induction:
For n=1, the LHS=RHS= $\frac{1}{1*2}$ . Let's assume it holds for n. Then, for n+1;
$\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{(n+1)(n+2 )}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=<br /> \frac{n(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}<br /> =\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{n+1}{n+2}$ And we're finally done

**Sillysidley** · 08-09-2007

Anyone with a new problem?

08-09-2007

Classic problem:

The coefficients of $x^5$ and $x^{15}$ in the expansion of $(1+x)^n$ are equal. Find n.

**Sillysidley** · 08-10-2007

We have nC5=nC15, so n=20

08-01-2007	#23
Sillysidley Join Date: Oct 2006 Posts: 822	Well, 2-9 can all be made (obviously) and you can get any any number higher by adding multiples of 10 (by adding 5s) so 1. Find $\frac{1}{12}+\frac{1}{23}+\frac{1}{34}+...+<br /> \frac {1}{499500}$ Oh, and if anyone cares, I meant to ask whats the largest number that must be a factor of any four consecutive odd numbers. (Three is too simple.) __________________ . Last edited by Sillysidley; 08-01-2007 at 08:52 PM..

08-01-2007	#25
Sillysidley Join Date: Oct 2006 Posts: 822	Looks like I had to go to the next line for some reason.. Oh, I forgot to metion, this isone of my favorite mathcount problems. It has two different approaches (One with a formula, one with a kind of series.) __________________ . Last edited by Sillysidley; 08-02-2007 at 05:35 PM..

08-04-2007	#26
Sillysidley Join Date: Oct 2006 Posts: 822	I'm gonna post the solution by 6:00 if no one answers by that time... __________________ .

08-04-2007	#27
Sillysidley Join Date: Oct 2006 Posts: 822	Eh, close enough The first one uses a telescoping series (basically where a bunch of things cancel out) and the second has a formula (and just to get better at proofs, I wanna prove it) Show more: We want all the terms except the first and last to cancel. So, we express it as: $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{499}-\frac{1}{500}=1-\frac{1}{500}=\frac{499}{500}$ 2nd: Show more: There's a formula that says the sum of n numbers of this kind sum up to $\frac{n}{n+1}$ . In other words, $\frac{1}{12}+\frac{1}{23}+...+\frac{1}{n(n+1)}= \frac{n}{n+1}$ We can then easily see the answer is $\frac{499}{500}$ . Proof: Easy enough to prove by induction: For n=1, the LHS=RHS= $\frac{1}{12}$ . Let's assume it holds for n. Then, for n+1; $\frac{1}{12}+\frac{1}{23}+...+\frac{1}{(n+1)(n+2 )}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=<br /> \frac{n(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}<br /> =\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{n+1}{n+2}$ And we're finally done __________________ . Last edited by Sillysidley; 08-04-2007 at 05:06 PM..

08-09-2007	#28
Sillysidley Join Date: Oct 2006 Posts: 822	Anyone with a new problem? __________________ .