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I have to prove that $\sum_{0}^{n}(-1)^i\binom{n}{i} = 0$,

I know i have to use the binomial theorem,But i dont know how.

Thanks.

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closed as off-topic by Daniel W. Farlow, Claude Leibovici, user147263, colormegone, John B Mar 16 '16 at 0:02

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  • $\begingroup$ Have you tried induction? $\endgroup$ – A. Prufrock Mar 15 '16 at 18:02
  • $\begingroup$ I have to prove it with binomial theorem. $\endgroup$ – Noam Mar 15 '16 at 18:03
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    $\begingroup$ Use the binomial theorem to expand $(1-1)^n$. $\endgroup$ – user296113 Mar 15 '16 at 18:03
  • $\begingroup$ @user296113 What do you mean ? $\endgroup$ – Noam Mar 15 '16 at 18:03
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    $\begingroup$ thanks, i solved it successfully using your help. $\endgroup$ – Noam Mar 15 '16 at 18:06
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$$ 0 = \bigl((1+ (-1)\bigr)^n = \sum_{i=0}^n \binom ni 1^{n-i}(-1)^i = \sum_{i=0}^n \binom ni (-1)^i $$

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The Binomial Theorem: $$(x+y)^n = \sum\limits_{i=0}^n \binom{n}{i}x^{n-i}y^i$$

For example $(x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4$

Using the above, notice what happens when $x=1$ and $y=-1$

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

In general, with other sums like this, one can throw in additional factors of $1$ to an arbitrary power to notice the pattern.

For example, $\sum\limits_{i=0}^n\binom{n}{i} = \sum\limits_{i=0}^n\binom{n}{i}(1)^i(1)^{n-i} = (1+1)^n = 2^n$

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