Prove that for every integer greater than 1. $\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$ How do you show the following.
Prove that for every integer greater than 1.
$\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$
My idea is that is that the whole sequence seem to alternate but it end as a negative.
So if the whole sequence adds up it will probably be negative one since it starts positive and ends negative.
And the the greatest integer function of (-1) is zero.
But I am not sure how to prove the following.
I did the binomial theorem for $(1-x)^n$ and got
$\dbinom{n}{0}1^{n}(-x)^{(0)}+\dbinom{n}{1}1^{n-1}(-x)^{(1)}+\dbinom{n}{2}1^{n-2}(-x)^{(2)}+\dbinom{n}{3}1^{n-3}(-x)^{(3)}+.....+\dbinom{n}{k}1^{n-k}(-x)^{(k)}$
but I find myself a bit stuck as the odds and even do not match.
then I got
$\dbinom{n}{0}1^{n}+\dbinom{n}{1}1^{n-1}(-x)+\dbinom{n}{2}1^{n-2}(x^2)+\dbinom{n}{3}1^{n-1}(-x^3)+\dbinom{n}{k} 1^{n-k}(x^k)$
ok I differentiated and I got $\dbinom{n}{1}-2\dbinom{n}{2}x+3\dbinom{n}{3}x^2-4x^3\dbinom{n}{4}+.....+\dbinom{n}{k} k x^{k-1}$

Hint: Expand $(1-x)^n$ using the Binomial Theorem and differentiate term by term.
Alternately, note that $k\binom{n}{k}=n\binom{n-1}{k-1}$.  
Details: For the first hint, let $n\gt 1$, and let $f(x)=(1-x)^n$. Then $f'(x)=n(1-x)^{n-1}$ and therefore $f'(1)=0$. 
We now compute $f'(x)$ another way, by first expanding $f(x)$ using the Binomial Theorem. We have 
$$f(x)=1-\binom{n}{1}x+\binom{n}{2}x^2 -\binom{n}{3}x^3 +\cdots +(-1)^n \binom{n}{n}x^n.$$
Differentiating, we get
$$f'(x)=-(1)\binom{n}{1}+(2)\binom{n}{2}x^1 -(3)\binom{n}{3}x^2+\cdots +(n)(-1)^n 
\binom{n}{n}x^{n-1}.$$
Put $x=1$. We get
$$f'(1)=-(1)\binom{n}{1}+(2)\binom{n}{2} -(3)\binom{n}{3}+\cdots +(n)(-1)^n 
\binom{n}{n}.$$
We already saw that the above expression is $0$ if $n\gt 1$. But the above expression is the negative of the expression you were asked to evaluate. 
The other hint gives another approach to the same problem. Using it, we find that 
$$\sum_1^n (-1)^{k-1}k \binom{n}{k}=n\sum_{1}^n(-1)^{k-1}\binom{n-1}{k-1}.$$ 
The expression on the right is equal to 
$$n\sum_{j=0}^{n-1}(-1)^j \binom{n-1}{j}.$$
Using the Binomial Theorem, we find this is $n(1-1)^{n-1}$, which is $0$. There are also more "combinatorial" ways of seeing that the sm of the binomial coefficients with alternating signs is $0$.  
