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Use the Binomial Theorem to Show

$$\sum_{i=0}^n (-1)^k \cdot C(n,k)$$

I'm not sure where to start here . . . I know it is missing an $a^{n-k}$ and a $b^{k}$ term that maybe I should set to be 1?

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  • $\begingroup$ Show what? This isn't a question. $\endgroup$ – Thomas Andrews Apr 21 '13 at 0:19
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HINT: You have the $b^k$ factor: it’s $(-1)^k$. The $a^{n-k}$ is invisible, because $a=\dots$?

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    $\begingroup$ Thanks! i get it now. So . . the answer is just as simple as $\sum_{i=0}^n C(n,k) \cdot (-1)^k \cdot 1^{n-k}$ ? $\endgroup$ – rbtLong Apr 21 '13 at 0:14
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    $\begingroup$ @rbtLong: That’s right, together with the fact that that sum is the binomial expansion of $\big(1+(-1)\big)^n$. $\endgroup$ – Brian M. Scott Apr 21 '13 at 0:18
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Yes. What is $0=(1+(-1))^n$ according to the binomial theorem?

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    $\begingroup$ +1. The rate at which some answers are given, I won't be surprised if an answer is posted to a question even before the question is asked. $\endgroup$ – user17762 Apr 21 '13 at 0:08
  • $\begingroup$ @user17762 I cannot claim I'm the fastest gun in SE, though! $\endgroup$ – Pedro Tamaroff Apr 21 '13 at 0:11
  • $\begingroup$ LOL yes you are faster than light itself. $\endgroup$ – rbtLong Apr 21 '13 at 0:12
  • $\begingroup$ @PeterTamaroff There are a lot of competitors (including you and me) for the fastest gun in SE! $\endgroup$ – user17762 Apr 21 '13 at 0:13
  • $\begingroup$ @user17762 I feel Brian will get accepted though... ¬¬ $\endgroup$ – Pedro Tamaroff Apr 21 '13 at 0:16

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