# No idea how to prove this :)

• Jun 9th 2010, 05:45 AM
hungthinh92
No idea how to prove this :)
$Give\ n\in N^*$

Prove:

$\sum_{k=0}^n \frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{1}{2}$
• Jun 9th 2010, 06:36 AM
p0oint
Just try to simplify $\frac{C_n^k}{C_{n+k+2}^{k+1}}$.

I came up with

$(n!(n+1)!)\sum_{k=0}^n \frac{(1+k)}{(n-k)!(2+k+n)!}$

Now just try to expand the sum, you'll notice something. :D
• Jun 9th 2010, 06:36 AM
Warrenx
Quote:

Originally Posted by hungthinh92
$Give\ n\in N^*$

Prove:

$\sum_{k=0}^n \frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{1}{2}$

Try L'Hopital's rule for the original equation
• Jun 9th 2010, 06:40 AM
Warrenx
Quote:

Originally Posted by p0oint
Just try to simplify $\frac{C_n^k}{C_{n+k+2}^{k+1}}$.

I came up with

$(n!(n+1)!)\sum_{k=0}^n \frac{(1+k)}{(n-k)!(2+k+n)!}$

Now just try to expand the sum, you'll notice something. :D

sigh I hate taylor series :(((
• Jun 9th 2010, 11:10 AM
hungthinh92
Quote:

Originally Posted by Warrenx
sigh I hate taylor series :(((

I hate it too
i got the answer. Po0int is awesome! But can you post the full answer?
• Jun 9th 2010, 01:43 PM
p0oint

$\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n}$

This can be found by using the binomial theorem $(1+x)^n$ at x=1.

I guess you need to use the binomial theorem somehow.

You can also proof by mathematical induction.

Also you may find useful these identities:

$\binom nk=\binom{n-1}{k-1}+\binom{n-1}k,\text{ for }0

$\binom nk = \binom n{n-k},\text{ for }0 \le k \le n.$

This task seems to me like some puzzle and tricky.

Try something and post it here.

Regards.
• Jun 10th 2010, 12:44 AM
CaptainBlack
Quote:

Originally Posted by Warrenx
Try L'Hopital's rule for the original equation

Since everything is discrete what do you think you might differentiate with respect to? Also this is not a limit problem.

CB
• Jun 10th 2010, 12:45 AM
CaptainBlack
Quote:

Originally Posted by Warrenx
sigh I hate taylor series :(((

That is not a Taylor series.

CB
• Jun 10th 2010, 05:04 AM
hungthinh92
Technically, nobody posted me the answer (Bow)
• Jun 10th 2010, 05:54 AM
Prove It
Quote:

Originally Posted by hungthinh92
Technically, nobody posted me the answer (Bow)

Technically, you have been given enough hints to be able to finish this problem off for yourself. Try it, and if you get stuck, post the work that you have done so that we can help you. We're not doing your work for you.
• Jun 10th 2010, 07:23 AM
hungthinh92
Quote:

Originally Posted by Prove It
Technically, you have been given enough hints to be able to finish this problem off for yourself. Try it, and if you get stuck, post the work that you have done so that we can help you. We're not doing your work for you.

I said
Quote:

Technically, nobody posted me the answer

Because I said

Quote:

i got the answer. Po0int is awesome!
Then Po0int said:

Quote:

This can be found by using the binomial theorem at x=1.

I guess you need to use the binomial theorem somehow.

You can also proof by mathematical induction.

Also you may find useful these identities:

This task seems to me like some puzzle and tricky.

Try something and post it here.

P/S: Prove It, I'm sorry but I just want somebody can show their work for this tricky problem. I found it after po0int's hint. If anybody needs the answer, I also can show my work. :(
• Jun 10th 2010, 12:05 PM
p0oint
@hungthinh92 please post your work here. Everybody is willing to help you, it's better for you to solve it by yourself. Trust me.

Regards.

P.S Mathematical induction seems very reasonable to me...
• Jun 10th 2010, 10:30 PM
simplependulum

$\sum_{k=0}^n \frac{ \binom{n}{k} }{ \binom{n+k+2}{k+1} }$

$=\sum_{k=0}^n \frac{ n! (k+1)! (n+k+2-k-1)!}{ k! (n-k)! (n+k+2)! }$

$= n!(n+1)! \sum_{k=0}^n \frac{ k+1 }{(n-k)! (n+k+2)!}$

Note that $(n-k) + (n+k+2) = 2n+2$ and we have

$\frac{n! (n+1)!}{(2n+2)!} \sum_{k=0}^n (k+1) \binom{2n+2}{n+k+2}$

Put $n + 1 = m$

the sum
$= \frac{(m-1)!m!}{(2m)! } \sum_{k=1}^m k \binom{2m}{m+k}$

$= \frac{(m-1)!m!}{(2m)! } \sum_{k=1}^m [(m+k)-m] \binom{2m}{m+k}$

$= \frac{(m-1)!m!}{(2m)! } \left[ \sum_{k=1}^m (m+k) \binom{2m}{m+k} - m \sum_{k=1}^m\binom{2m}{m+k} \right]$

Since $\binom{2m}{m+k} = \frac{2m}{m+k} \binom{2m-1}{m+k-1}$

please check it , using the symmetrical property of binomial coefficients

$\sum_{k=1}^m (m+k) \binom{2m}{m+k} = 2m \sum_{k=1}^m \binom{2m-1}{m+k-1} = (2m) \frac{ 2^{2m-1}}{2} = m 2^{2m-1}$

and

$\sum_{k=1}^m\binom{2m}{m+k} = \frac{ 2^{2m} - \binom{2m}{m} }{2} = 2^{2m-1}- \frac{\binom{2m}{m}}{2}$

so we have the sum

$= \frac{(m-1)!m!}{(2m)! } [m 2^{2m-1} - m2^{2m-1} + m\frac{\binom{2m}{m}}{2}] = \frac{(m!)^2}{(2m)! } \frac{\binom{2m}{m}}{2}$

$= \frac{1}{ \binom{2m}{m}} \frac{\binom{2m}{m}}{2} = \frac{1}{2}$
• Jun 11th 2010, 01:34 AM
p0oint
@simplependulum you should not do that. He should solve that by himself.

By math. induction you can prove this in 5 min.

At least he should try to do something.

Also there is a better solution and faster which gaves you:

$
= \sum_{k=0}^n \frac{ k+1 }{(n-k)! (n+k+2)!} = \frac{2+n}{2n!(2+n)!}
$

Then

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

Regards.
• Jun 11th 2010, 03:21 AM
undefined
Quote:

Originally Posted by p0oint
@simplependulum you should not do that. He should solve that by himself.

I generally try to allow people to solve their own problems, but I'm not against posting full solutions, following the reasoning of this post (scroll down to the "rant"). So, I don't think simplependulum should be scolded, although I see where you're coming from, the OP hasn't put any effort into typing out work.