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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.

Thanks in advance !

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

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Hint;

$$i\binom{n}{i}=n\binom{n-1}{i-1}$$

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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 $ $n>1,n>-1,n=1$

Thanks for help !

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