Algebric identity prove: sum[m=0,n] (m-k)/(n-r)*(C(m,k) C(n-m,r-k))/C(n+1,r+1) = ...

[paul2834]

Hi,

I need to prove the following identity:

. . . . .\(\displaystyle \large \displaystyle \sum_{m = 0}^n\, \dfrac{m\, -\, k}{n\, -\, r}\, \cdot\, \dfrac{\binom{m}{k}\, \binom{n - m}{r - k}}{\binom{n+1}{r+1}}\, =\, \dfrac{k\, +\, 1}{r\, +\, 2}\)

I tried to go straight forward with the definition of (n choose k), but I got stuck. Do you know how can I prove this?

 

[Deleted member 4993]

I need to prove the following identity:

. . . . .\(\displaystyle \large \displaystyle \sum_{m = 0}^n\, \dfrac{m\, -\, k}{n\, -\, r}\, \cdot\, \dfrac{\binom{m}{k}\, \binom{n - m}{r - k}}{\binom{n+1}{r+1}}\, =\, \dfrac{k\, +\, 1}{r\, +\, 2}\)

I tried to go straight forward with the definition of (n choose k), but I got stuck. Do you know how can I prove this?

That is the way to do this problem. Where did you get stuck? Please share your work - so that we can unstuck you.

 

[paul2834]

First, I forgot to mention that k < r in this question. Now, we can see that for each m such that 0 < m < k we are summing zeros, so we consider the sum from m = k + 1 to m = n.

Now, I tried to see what I get at m = k + 1 and at m = n, because I thought I might get k + 1 at one expression and 1/(r + 2) at the other expression or something like that (and the other expressions in the sum might cancel each other)- but I didn't get it.

This is what I got for m = k + 1:

. . .\(\displaystyle \large \displaystyle \dfrac{1}{n\, -\, r}\, \cdot\, \dfrac{(k\, +\, 1)\, \binom{n-k-1}{r-k}}{\binom{n+1}{r+1}}\)

. . . . .\(\displaystyle \large \displaystyle =\, \dfrac{(k\, +\, 1)\, (n\, -\, k\, -\, 1)!}{(r\, -\, k)!\, (n\, -\, r\, -\, 1)!}\, \cdot\, \dfrac{(r\, +\, 1)\, (n\, -\, r\, -\, 1)!}{(n\, +\, 1)!}\)

. . . . .\(\displaystyle \large \displaystyle =\, (k\, +\, 1)\, \cdot\, \dfrac{(r\, +\, 1)\, \cdot\, ...\, \cdot\, (r\, -\, k\, +\, 1)}{(n\, +\, 1)\, \cdot\, ...\, \cdot\, (n\, -\, k)}\)

And for m = n:

. . .\(\displaystyle \large \displaystyle \dfrac{n\, -\, k}{n\, -\, r}\, \cdot\, \dfrac{\binom{n}{k}}{\binom{n+1}{r+1}}\)

. . . . .\(\displaystyle \large \displaystyle =\, \dfrac{n!}{k!\, (n\, -\, k\, -\, 1)!}\, \cdot\, \dfrac{(r\, +\, 1)!\, (n\, -\, r\, -\, 1)!}{(n\, +\, 1)!}\)

. . . . .\(\displaystyle \large \displaystyle =\, \dfrac{(r\, +\, 1)!\, (n\, -\, r\, -\, 1)\, \cdot\, ...\, \cdot\, (n\, -\, k)}{k!\, (n\, +\, 1)}\)