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I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$.

Marc van Leeuwen
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JSchlather
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    is the proof you looking for using $(1+1)^n=2^n$? – jimjim Jan 24 '11 at 04:30
  • Well, no. That one I was also aware of. It's more of a curiosity if there's any direct method to go from the summation to $2^n$. – JSchlather Jan 24 '11 at 04:40
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    One should not think of the algebraic and combinatorial proofs as different. There is a straightforward dictionary between algebra and combinatorics in these cases (and it is given by taking generating functions). – Qiaochu Yuan Jan 24 '11 at 09:12
  • Zeilberger's algorithm might do it - it's a useful tool for this kind of problem in general (sum from $-\infty$ to $\infty$ of a hypergeometric with finite support). – Peter Taylor Jan 24 '11 at 09:29
  • @Peter Taylor: Zeilberger's algorithm produces the recurrence given in my answer. See Section 5.8 of Concrete Mathematics. – Mike Spivey Jan 24 '11 at 14:15
  • I don't quite understand the question for a proof. Isn't $\small (a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1}b + \ldots + \binom{n}{n} b^n $ just the definition of the binomials (at least historically)? Then - what might be the point of proving ? – Gottfried Helms Dec 04 '11 at 23:05
  • upps - just overlooked that this thread is a fairly old hat. sorry... – Gottfried Helms Dec 04 '11 at 23:12
  • @JacobSchlather What would be more direct than $(1+1)^n=2^n$? – Pedro Apr 27 '12 at 18:08
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    @PeterTamaroff It was a sort of silly question that I asked over a year ago because I had always thought of binomial coefficients being defined as $n!/(k!(n-k)!)$ and was always curious how one could go from this sum of factorials divided by other factorials to $2^n$. When I had said direct, I had meant something along the lines of proving the binomial recurrence by moving factorials around. It's not a very clear question in retrospect. But people seemed to get the basic idea of what I wanted. – JSchlather Apr 27 '12 at 19:23

5 Answers5

25

Here's one. Let $g(n) = \sum \limits_{i=0}^n \binom{n}{i}$. Then

$$g(n+1) - g(n) = \sum_{i=0}^{n+1} \binom{n+1}{i} - \sum_{i=0}^n \binom{n}{i} = \sum_{i=0}^{n+1} \left(\binom{n+1}{i} - \binom{n}{i}\right) = \sum_{i=0}^{n+1} \binom{n}{i-1} $$ $$= \sum_{i=0}^n \binom{n}{i} = g(n).$$ Here, we use the fact that $\binom{n}{n+1} = \binom{n}{-1} = 0$, as well as the binomial recurrence $\binom{n+1}{i} = \binom{n}{i} + \binom{n}{i-1}$.

Thus we have $g(n+1) = 2g(n)$, with $g(0) = 1$. Since $g(n)$ doubles each time $n$ is incremented by 1, we must have $$g(n) = \sum_{i=0}^n \binom{n}{i} = 2^n.$$

Quixotic
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Mike Spivey
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19

Simply use the binomial formula.

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

With $a = b = 1$ you have your result.

Mitch
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AbstractDissonance
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Well, here is one.

$$\sum_{i=0}^n \binom{n}{i}=2^n$$ $$\sum_{i=0}^n \binom{n}{i}+\sum_{i=0}^n \binom{n}{i}=2^{n+1}$$ $$\binom{n}{0}+\left [ \binom{n}{0}+\binom{n}{1} \right ]+...+\left [ \binom{n}{n-1}+\binom{n}{n}\right ]+\binom{n}{n}=2^{n+1}$$ $$\sum_{i=0}^{n+1} \binom{n+1}{i}=2^{n+1}$$

picakhu
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3

Using the binomial theorem, we find:

$2^n = (1 + 1)^n = \sum\limits_{k=0}^{n} \binom{n}{k} 1^k*1^{n-k} = \sum\limits_{k=0}^{n} \binom{n}{k}$

3

You could use exponential generating functions to prove this identity. $$\begin{eqnarray}\sum_{n\ge0}2^n\frac{x^n}{n!}&=&\sum_{n\ge0}\frac{(2x)^n}{n!}\\&=&e^{2x}\\&=&e^xe^x\\&=&\sum_{i\ge0}\frac{x^i}{i!}\sum_{j\ge0}\frac{x^j}{j!}\\&=&\sum_{n\ge0}\sum_{i=0}^{n}\frac{x^i}{i!}\frac{x^{n-i}}{(n-i)!}\\&=&\sum_{n\ge0}\sum_{i=0}^n\binom{n}{i}\frac{x^n}{n!}\end{eqnarray} $$ Now, comparing the coefficients of $x^n$ for ${n\ge0}$ gives the identity.

Markus Scheuer
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