1. ## Show

Show that $(\frac{1}{2})[\frac{(-3)(-4)...(-3-n+1)}{n!}(-\frac{1}{2})^n]=(\frac{1}{2})^{n+2}(n+1)(n+2)$

2. ## Re: Show

start taking -1 out from the left side

$(-1)^n\frac{1}{2}\left[\frac{(3)(4)\cdots(3+n-1)}{n(n-1)\cdots 2.1}\right]\left[-\frac{1}{2}\right]^n$

since there are total n terms in the numerator, we get a factor of $(-1)^n$
out

$(-1)^{2n}\left[\frac{1}{2}\right]^{n+1}\frac{(3)(4)\cdots(n+2)}{n(n-1)\cdots 4.3.2.1}\right]$

$((-1)^2)^n\left[\frac{1}{2}\right]^{n+1}\frac{(3)(4)\cdots n(n+1)(n+2)}{n(n-1)\cdots 4.3.2.1}\right]$

now do the cancellations

$((-1)^2)^n\left[\frac{1}{2}\right]^{n+1}\frac{(n+1)(n+2)}{2.1}\right]$

$\left[\frac{1}{2}\right]^{n+2}(n+1)(n+2)$