[jacks]

The sum \( \displaystyle \sum^{n}_{k=1}\frac{\binom{n+k}{k}}{\binom{n}{k}}\frac{2k-1}{n+k}= \)

[Richard R Richard]

https://www.wolframalpha.com/input?i2d=true&i=Sum%5B%5Cfrac%5C%28123%29%5Cbinom%5C%28123%29n%2Bk%5C%28125%29%5C%28123%29k%5C%28125%29%5C%28125%29%5C%28123%29%5Cbinom%5C%28123%29n%5C%28125%29%5C%28123%29k%5C%28125%29%5C%28125%29%5Cfrac%5C%28123%292k-1%5C%28125%29%5C%28123%29n%2Bk%5C%28125%29%2C%7Bk%2C1%2Cn%7D%5D&lang=es


It does not have a very simplified solution

[Luis Fuentes]

Hola

It does not have a very simplified solution

What about this?

\( \displaystyle \sum^{n}_{k=1}\frac{\binom{n+k}{k}}{\binom{n}{k}}\frac{2k-1}{n+k}=\displaystyle\binom{2n}{n}-1 \)

But, for now I don't know how to prove it

Best regards:

[Richard R Richard]

I try

$$\dfrac{\binom{n+k}{k}}{\binom{n}{k}}=$$$$\dfrac{\dfrac{(n+k)!}{n!\cancel{k!}}}{\dfrac{n!}{(n-k)!\cancel{k!}}}=$$$$\dfrac{(n+k)!(n-k)!}{n!^2}=\dfrac{\left[\prod\limits_{i=n-k+1}^{n+k}i\right]\cancel{(n-k)!^2}}{\left[\prod\limits_{i=n-k+1}^{n}i\right]^2\cancel{(n-k)!^2}}=$$$$\dfrac{\prod\limits_{i=n-k+1}^{n+k}i}{\left[\prod\limits_{i=n-k+1}^{n}i\right]^2}=$$$$\dfrac{\prod\limits_{i=n+1}^{n+k}i}{\prod\limits_{i=n-k+1}^{n}i}=$$

$$\displaystyle \sum^{n}_{k=1}\left[\dfrac{\prod\limits_{i=n+1}^{n+k}i}{\prod\limits_{i=n-k+1}^{n}i}\right]\frac{2k-1}{n+k}=$$$$\displaystyle \sum^{n}_{k=1}\left[\dfrac{\prod\limits_{i=n+1}^{n+k-1}i}{\prod\limits_{i=n-k+1}^{n}i}\right](2k-1)=$$$$\displaystyle\binom{2n}{n}-1 $$ ? I don't know Rick....