1. ## Binomial Formula

Prove, that for every integer n>1,

$\displaystyle \binom{n}{1}$ - 2$\displaystyle \binom{n}{2}$ +3$\displaystyle \binom{n}{3}$ +...+$\displaystyle (-1)^{n-1}$n$\displaystyle \binom{n}{n}$=0

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

The teacher gave us a hint that said we should differentiate and then plug in x =-1. I'm having a hard time converting this to the binomial theorem and understand why x=-1 shows it for all 'n'

2. $\displaystyle (1+x)^n = 1 + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \binom{n}{4} x^4 ..... \binom{n}{n} x^n$

differentiate both sides with respect to x

$\displaystyle n(1+x)^{n-1} = \binom{n}{1} + 2 \binom{n}{2} x +3 \binom{n}{3} x^2 + 4 \binom{n}{4} x^3 ..... n \binom{n}{n} x^{n-1}$

sub x = -1

$\displaystyle n(1-1)^{n-1} = \binom{n}{1} - 2 \binom{n}{2} +3 \binom{n}{3} - 4 \binom{n}{4} ..... + (-1)^{n-1} n \binom{n}{n}$

$\displaystyle 0 = \binom{n}{1} - 2 \binom{n}{2} +3 \binom{n}{3} - 4 \binom{n}{4} ..... + (-1)^{n-1} n \binom{n}{n}$

3. Just a little typo: $\displaystyle (-1)^{n-1}$ twice...