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I have noticed that two of the math books used in high school goes about the binomial theorem in this way:

  1. Prove it on integers using induction
  2. Generalize it and use it in proofs needed to develop calculus
  3. Use calculus to prove the binomial theorem for $\mathbb{R}$

In college level text books (Rudin...) this approach is of cause not taken, but these are often to advanced to show in high school.

There is a rather simple outline here that develops most of the foundation needed for proving the binomial theorem for $\mathbb{R}$. But I was wondering if anyone knows of a simple proof suitable to showing in a high school class for interested although not advanced students ?.

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    @ArturoMagidin I see your $p$oints, it was not me intention to come of that strongly and as a non native english person I apologize for the spelling mistakes. As for the question I was not thinking about graduate level math where the proofs are not circular (i.e. Rudin's books and the likes) but the kind of text books used for calculus in high school which (at least in Denmark) does have this problem (among many other ones such as the description of continuous functions).2012-02-02

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I don't fully understand what you are asking. In any case, here is a different way to prove the theorem, it depends heavily on the Gamma function, and particular integrals. I doubt it is what you want, but it may be useful to someone:

We want to prove the generalized binomial theorem, namely that $(1+x)^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}$where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$ Since increasing $\alpha$ by an integer amount is simply multiplying by $(1+x)^{n}$, and is easily dealt with, we need only consider some interval of length $1$. Suppose $\alpha\in[-1,0).$ We can write $\binom{\alpha}{k}=\frac{(-1)^{k}(-\alpha)(1-\alpha)\cdots(k-1-\alpha)}{k!}.$Multiplying the top and bottom by $\Gamma\left(-\alpha\right),$and using the properties of Gamma gives $\binom{\alpha}{k}=\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!\Gamma(-\alpha)}.$ Recall that for $s>0$, $\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt.$ Then $\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}=\frac{1}{\Gamma(-\alpha)}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!}$ and by using this definition of $\Gamma(s)$, and switching the order we have

$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t}t^{-\alpha-1}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}t^{k}}{k!}dt.$Recognizing the series, we have$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t(1+x)}t^{-\alpha-1}dt.$Substituting $u=t(1+x)$, this becomes $(1+x)^{\alpha}\frac{\int_{0}^{\infty}e^{-u}u^{-\alpha-1}du}{\Gamma(-\alpha)}=(1+x)^{\alpha},$which was the desired result.

Maybe that helps,

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    @Qiaochu: Dear Qiaochu, Did you mean "*not* well-posed"? Regards,2011-06-16