$\displaystyle Give\ n\in N^*$
Prove:
$\displaystyle \sum_{k=0}^n \frac{C_n^k}{C_{n+k+2}^{k+1}}=\frac{1}{2}$
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.
I saidTechnically, nobody posted me the answer
Because I said
Then Po0int said:i got the answer. Po0int is awesome!
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.
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} $
@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.
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.