$\displaystyle Give\ n\in N^*$

Prove:

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

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- Jun 9th 2010, 04:45 AMhungthinh92No idea how to prove this :)
$\displaystyle Give\ n\in N^*$

Prove:

$\displaystyle \sum_{k=0}^n \frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{1}{2}$ - Jun 9th 2010, 05:36 AMp0oint
Just try to simplify $\displaystyle \frac{C_n^k}{C_{n+k+2}^{k+1}}$.

I came up with

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

Now just try to expand the sum, you'll notice something. :D - Jun 9th 2010, 05:36 AMWarrenx
- Jun 9th 2010, 05:40 AMWarrenx
- Jun 9th 2010, 10:10 AMhungthinh92
- Jun 9th 2010, 12:43 PMp0oint
I will give you few tips that may help you:

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

This can be found by using the binomial theorem $\displaystyle (1+x)^n$ 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:

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

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

This task seems to me like some puzzle and tricky.

Try something and post it here.

Regards. - Jun 9th 2010, 11:44 PMCaptainBlack
- Jun 9th 2010, 11:45 PMCaptainBlack
- Jun 10th 2010, 04:04 AMhungthinh92
Technically, nobody posted me the answer (Bow)

- Jun 10th 2010, 04:54 AMProve It
- Jun 10th 2010, 06:23 AMhungthinh92

I saidQuote:

Technically, nobody posted me the answer

Because I said

Quote:

i got the answer. Po0int is awesome!

Quote:

I will give you few tips that may help you:

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, 11:05 AMp0oint
@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, 09:30 PMsimplependulum
Your series should be :

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

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

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

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

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

Put $\displaystyle n + 1 = m $

the sum

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

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

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

please check it , using the symmetrical property of binomial coefficients

$\displaystyle \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

$\displaystyle \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

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

$\displaystyle = \frac{1}{ \binom{2m}{m}} \frac{\binom{2m}{m}}{2} = \frac{1}{2} $ - Jun 11th 2010, 12:34 AMp0oint
@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:

$\displaystyle

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

$

Then

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

Regards. - Jun 11th 2010, 02:21 AMundefined
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.