Problem 7 fall 2011

As I said problem 7 is much easier than problem 6. When I was on the bus, I made a mental draft procedure: the binomail theorom. Using the expansion of it.

The problem says: For every $n\geq2 $, $$\sum_{k=1}^n \dbinom{n}{k}k (-1)^k =0$$. where $\dbinom{n}{k}$ is the usual binomial coefficient.

pf:
Let $$ f(x)=(x-1)^n= \sum_{k=0}^n \dbinom{n}{k} x^k (-1)^{n-k}$$


then $$ \frac{d}{dx}f(x) = n(x-1)^{n-1} = \sum_{k=1}^n \dbinom{n}{k}kx^{k-1} (-1)^{n-k}$$.

It is know that $x=1$ is the $n-1$ degree root of $ \frac{d}{dx}f(x) = 0$.

consider n is even number, then $(-1)^{n-k}=(-1)^k$, for k not greater than n.

Then,$\frac{d}{dx}f(x) \Big|_{x=1}=\sum_{k=1}^n \dbinom{n}{k} k (-1)^k =0 $

If $n$ is odd number, then .  
[Rewise Here]. When I was tutoring a student today, I found there is a problem in last term of the original expansion. $\dbinom{n}{n}kx^{k-}(-1)^{n-n-1}$. Maybe that will ok, but I multiple $1=(-1)^2$ to the $\frac{d}{dx}f(x)$. 


Because $(-1)^2=1$
$$(-1)^2 \sum_{k=1}^n \dbinom{n}{k} k x^{k-1} (-1)^{n-k} = (-1)\sum_{k=1}^n \dbinom{n}{k} k x^{k-1}(-1)^{n-k+1}$$
Notice that $(-1)^{n-k+1}=(-1)^k $.
Then $ \frac{d}{dx}f(x) \Big|_{x=1} =(-1)\sum_{k=1}^n \dbinom{n}{k} k (-1)^k =0 $ That implies  
$$\sum_{k=1}^n \dbinom{n}{k} k (-1)^{n-k} =0 $$ is hold for all the $n\geq2 $.

To complete the proof, there should add the condition n >=2 somewhere, which implies that 1 is the n-degree  root of f(x), and  (n-1)-degree root of f'(x)=0. Because n>=2,  n-1>=1.

Maybe there is typing error. But the steps make sense.  If so, the scale of the difficulty of the problem set it too large.