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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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closed as not constructive by Qiaochu Yuan Jun 17 '11 at 10:06

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

In my experience, the instances of the binomial theorem used "to develop calculus" all involve integer (in fact, usually positive integer) exponents, so that "circularity" would not enter into it... – Arturo Magidin Jun 16 '11 at 21:22
What does $x^y$ mean without calculus? – jspecter Jun 16 '11 at 21:28
As noted, question is not clear. Should be closed unless Lars returns and improves it. – GEdgar Jun 16 '11 at 21:56
See my answer here. – Bill Dubuque Jun 16 '11 at 21:58
Despite having taught some version of calculus for too many years, I cannot think of anything in the development that needs to rely on the Binomial Theorem, even for positive integers. For example, the derivative of $x^n$ can be found by factoring $x^n-a^n$, or by induction. (If one uses induction, it is best for weaker classes not to bring out the formal machinery explicitly.) Newton's heuristic discovery of a version of the general Binomial Theorem did play a significant role in his initial development of the calculus, but we need not, and generally do not, imitate him. – André Nicolas Jun 16 '11 at 22:39

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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How is this an answer to your question? Eric is using results from calculus that surely depend on infinite series such as Riemann sums. (I am not saying that your question is well-posed, only that this doesn't seem to be an answer to it.) – Qiaochu Yuan Jun 16 '11 at 22:12
@Qiaochu: Dear Qiaochu, Did you mean "not well-posed"? Regards, – Matt E Jun 16 '11 at 23:44

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