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so I have to prove this using the binomial theorem: $$\sum_{k=0}^n{(-1)^k\begin{pmatrix}n\\k\end{pmatrix}}=0$$

I know the binomial theorem states: $(x+y)^n=\sum_{k=0}^n{\begin{pmatrix}n\\k\end{pmatrix}x^{n-k}\cdot y^k}$.

I can also see that the $(-1)^k$ will be alternating so I will get $+1-1+1-1...$. However, I'm not sure how to show it will $=0$. Thanks.

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You can use the "twist" of your binomial theorem $$(x-y)^n=\sum_{k=0}^n (-1)^k\binom{n}{k}x^{n-k}y^k.$$

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  • $\begingroup$ This will prove that it will = 0? $\endgroup$ – KFC Sep 25 '15 at 11:54
  • $\begingroup$ @KFC you only need to let $x=y=1$. $\endgroup$ – Asydot Sep 25 '15 at 12:02

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