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I want to know a simple way to prove $$\sum_{k=0}^n {{n}\choose{k}} = 2^n.$$ Our teacher only gave us the combinatorial proof.

gt6989b
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ydnfmew
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    This question on cross validated. – Brian61354270 Mar 29 '19 at 01:56
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    IMHO, the combinatorial way IS the simple way. The other fun one is to expand $(x+y)^n$ and toss in $x=y=1$. – Randall Mar 29 '19 at 01:56
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    Personally, I don't feel the combinatorial way is necessarily the "simpler" way, depending on your exposure. Though I will agree it's at least the more insightful and interesting way. – PrincessEev Mar 29 '19 at 02:00
  • If I were sitting in on your class, I would shake my finger at your teacher for not showing you the easy proof, as given by @EeveeTrainer below. – Lubin Mar 29 '19 at 02:04
  • @Lubin I might disagree. If you need to prove the BT first, it's much more annoying and gross. – Randall Mar 29 '19 at 02:06
  • See also https://math.stackexchange.com/questions/2092320/alternative-proof-that-sum-j-0n-binomnj-binomn0-cdots-bino?rq=1 – Mike Earnest Mar 29 '19 at 04:12
  • Well, @Randall, I just ran through the proof of BT, and it’s perfectly easy, if you use the recursive definition of $\binom nk$ rather than the one with factorials. – Lubin Mar 29 '19 at 04:29
  • See also https://math.stackexchange.com/questions/519832/proving-by-induction-that-sum-k-0nn-choose-k-2n – Mike Earnest Mar 30 '19 at 04:36

1 Answers1

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Let's assume the binomial theorem can be taken for granted. By it, for $n \in \Bbb N$,

$$(a+b)^n = \sum_{k=0}^n \binom n k a^k b^{n-k}$$

Take $a=b=1$. Then we have

$$(1+1)^n = \color{blue}{ 2^n = \sum_{k=0}^n \binom n k } = \sum_{k=0}^n \binom n k 1^k 1^{n-k}$$

since $1$ to any power is just $1$.

PrincessEev
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  • Just to note, the simplest proof of the binomial theorem (admittedly subjective, as you point out in your comment) for positive integral powers is probably the combinatorial one: The coefficient of $a^{k}b^{n - k}$ in the expansion of $(a + b)^n$ is the number of ways of choosing $k$ $a$'s and $n - k$ $b$'s from the $n$ brackets $(a + b)$, which is the number of ways of choosing $k$ brackets out of $n$, from which the $a$'s can be chosen, which is $\binom n k$. – M. Vinay Mar 29 '19 at 02:49
  • Gosh, @M.Vinay, I would define $\binom nk$ recursively, right outta the Pascal triangle. The proof drops out that way too. – Lubin Mar 29 '19 at 04:32