1. ## Binomial Theorem?

Prove that:

(n!m!)/(m + n)! = (nC1)/(m + 1) - 2.(nC2)/(m + 2) + 3.(nC3)/(m + 3) - ... + (-1)^(n+1)n.(nCn)/(m + n)

NOTE: nC1 means n choose 1. I struggled with latex so I had to eventually use this notation.

2. This question seems similar:

Find a closed form for the expression

for .

.

But I am unable to form the required integral in my problem.

3. Yes you are right this is the same kind of problem

$\displaystyle \sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \sum_{k=1}^{n} \binom{n}{k} \int_{0}^{1} k\-1)^{k-1} x^{m+k-1} dx$

$\displaystyle \sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-1)^{k-1} x^{k-1} dx$

$\displaystyle \sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} dx$

Now
$\displaystyle \sum_{k=0}^{n} \binom{n}{k} (-x)^{k} = (1-x)^n$

By differentiation with respect to x
$\displaystyle \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} = n\1-x)^{n-1}$

Therefore
$\displaystyle \sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \:n\1-x)^{n-1}dx$

$\displaystyle \sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = n\:\int_{0}^{1} x^m \1-x)^{n-1}dx$

Now you just need to show that
$\displaystyle n\:\int_{0}^{1} x^m \1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}$