I have problems prooving following equation. I tried to use induction, but I did not suceed.
$$\sum_{k=0}^{\lfloor \frac {n}{2} \rfloor}\binom{n}{k} \frac{n-2k+1}{n-k+1}= \binom{n}{\lfloor \frac {n}{2} \rfloor}$$
The context of the equation is combinatorical:
Let $\lambda$ be a partition of n, then $f^\lambda$ ist the amount of Young Tableauxs of shape $\lambda$. (See also: https://en.wikipedia.org/wiki/Young_tableau#Tableaux)
I found out using the hook-formula: $f^{(n-k,k)}=\binom{n}{k} \frac{n-2k+1}{n-k+1}$. So the upper equation I am looking for is the sum over these amounts of Young-Tableauxs.
Thanks in advance
We have: $$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n \choose k} \frac{n-2k+1}{n-k+1}=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n \choose k} \bigg(1-\frac{k}{n-k+1}\bigg)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n \choose k} -\frac{n!}{(k-1)!(n-k+1)!}$$ But we know that $\frac{n!}{(k-1)!(n-k+1)!}={n \choose {k-1}}$. So- $$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n \choose k} \frac{n-2k+1}{n-k+1}=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n \choose k} -{n \choose {k-1}}={n \choose {\lfloor \frac{n}{2} \rfloor}}$$ due to telescoping.