# Binomial coefficients-sums

I need help solving this task so if anyone had a similar problem it would help me.

Calculate:

$$\sum\limits_{i=0}^{n}(-1)^i i {n \choose i}$$

I tried this:

$$\sum\limits_{i=0}^{n}(-1)^i i \frac{n!}{i!(n-i)!}\\\sum\limits_{i=0}^{n}(-1)^i \frac{n!}{(i-1)!(n-i)!}\\\sum\limits_{i=0}^{n}\frac{(-1)^i n!}{(i-1)!(n-i)!}$$

And now with this part I don’t know what to do next.

• You do not want to be thinking about $(i-1)!$ when $i=0$ – Henry Sep 15 at 8:35

Let $$f(x)=\sum\limits_{i=0}^{n} \binom {n} {i} (-x)^{i}$$. By Binomial theorem $$f(x)=(1-x)^{n}$$. Also $$f'(1)=\sum\limits_{i=0}^{n} (-1)^{i} i\binom {n} {i}$$. Hence the answer is $$f'(1)=n(1-x)^{n-1}(-1)|_{x=1}=0$$ if $$n >1$$ and$$-1$$ if $$n=1$$.
$$i\binom{n}{i}=n\binom{n-1}{i-1}$$
I did it like this, so I'm looking for your thoughts, is it right? $$\sum\limits_{i=0}^{n} (-1)^{i} i\binom {n} {i}\\\sum\limits_{i=0}^{n} (-1)^{i} n\binom {n-1} {i-1}\\\sum\limits_{i=0}^{n} (-1)^{i} n\frac{(n-1)!}{(i-1)!(n-i)!} \\\sum\limits_{i=0}^{n} (-1)^{i} n(n-1) \frac{1}{(i-1)!(n-i)!}\\n(n-1)!\sum\limits_{i=0}^{n} (-1)^{i}\frac{1}{(i-1)!(n-i)!}\\ n\sum\limits_{i=0}^{n} (-1)^{i}\frac{(n-1)!}{(i-1)!(n-i)!}\\n\sum\limits_{i=0}^{n}\binom{n-1}{i-1}(-1)^{i}\\n(1-x)^{n-1}=0$$ $$n>1,n>-1,n=1$$