No idea how to prove this :)

**hungthinh92** · June 9th 2010, 04:45 AM

$Give\ n\in N^*$

Prove:

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

**p0oint** · June 9th 2010, 05:36 AM

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.

**Warrenx** · June 9th 2010, 05:36 AM

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

**Warrenx** · June 9th 2010, 05:40 AM

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.

sigh I hate taylor series

((

**hungthinh92** · June 9th 2010, 10:10 AM

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?

**p0oint** · June 9th 2010, 12:43 PM

I will give you few tips that may help you:

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

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:

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

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

This task seems to me like some puzzle and tricky.

Try something and post it here.

Regards.

**CaptainBlack** · June 9th 2010, 11:44 PM

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

**CaptainBlack** · June 9th 2010, 11:45 PM

Originally Posted by Warrenx

sigh I hate taylor series

((

That is not a Taylor series.

CB

**hungthinh92** · June 10th 2010, 04:04 AM

Technically, nobody posted me the answer

**Prove It** · June 10th 2010, 04:54 AM

Originally Posted by hungthinh92

Technically, nobody posted me the answer

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.

**hungthinh92** · June 10th 2010, 06:23 AM

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

Technically, nobody posted me the answer

Because I said

i got the answer. Po0int is awesome!

Then Po0int said:

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.

**p0oint** · June 10th 2010, 11:05 AM

@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...

**simplependulum** · June 10th 2010, 09:30 PM

Your series should be :

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

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

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

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

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

$= \frac{1}{ \binom{2m}{m}} \frac{\binom{2m}{m}}{2} = \frac{1}{2}$

**p0oint** · June 11th 2010, 12:34 AM

@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:

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

Then

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

Regards.

**undefined** · June 11th 2010, 02:21 AM

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.

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