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I found the exact same question that I want to ask here : Proving $\sum\limits_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$

From my understanding this doesn't use Inclusion Exclusion Principle and if it does in some way, it was answered a little too briefly for me to comprehend it, i am still new to discrete mathematics and have trouble understanding the concepts.

I am trying to prove the following:

$$\sum_{k=0}^n (-1)^k {n \choose k} = 0$$

How would I use IEP to prove the following?

AFC
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1 Answers1

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Inclusion-Exclusion principle: $$ \left|\,\bigcup_{i=1}^n A_i\,\right|=\sum_{k=1}^n(-1)^{k-1} \sum_{1\leq i_1<i_2<\ldots<i_k\leq n} \left|A_{i_1}\cap A_{i_2}\cap \ldots \cap A_{i_k}\right|. $$

So take $A_i=\{1\},\,\forall i=1,\cdots,n$, and the result follows.

Hope this helps.

awllower
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