From trial and error I know that $$\sum_{k=0}^n\binom{n}{k}= 2^n$$ Remember studying it. But can't prove it theoretically now. Can someone prove the above
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*Hint:* Do you know that $(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$? – Devansh Kamra Aug 27 '20 at 15:38
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You might want to look here: https://math.stackexchange.com/questions/519832/proving-by-induction-that-sum-k-0nn-choose-k-2n – Aiden Chow Aug 27 '20 at 15:42
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3Does this answer your question? Proving by induction that $ \sum_{k=0}^n{n \choose k} = 2^n$ – Devansh Kamra Aug 27 '20 at 15:43
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Another variation: https://math.stackexchange.com/questions/137727/evaluate-sum-limits-k-0n-binomnk-combinatorially – David K Aug 27 '20 at 16:34
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$$\sum_{k=0}^n\binom{n}{k}=\sum_{k=0}^n\binom{n}{k} 1^k 1^{n-k} = (1+1)^n= 2^n$$
TheSilverDoe
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Counting up every way you choose anything from $n$ things, is the same as for each thing you choose whether to include it or not.
cr001
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