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It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:

$ (x+y)^n=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r $

Unfortunately, I wasn't able to find any clear explanation of how to get from the regular theorem to the generalised one, the only proof I found being based on some obscure mathematics, while my math book entirely skips the explanation.

Hence, my question is how do you prove the generalised theorem by deriving it from the regular theorem or otherwise?

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    "why don't they teach it in secondary school?" - That's what math.SE is for.2014-06-06

3 Answers 3

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Here is another approach: The formula $(1+x)^\alpha=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(1)$ is true for $\alpha\in{\mathbb N}$; so maybe its true for arbitrary $\alpha$. To find out fix an $\alpha\in{\mathbb R}$ and consider the function $f(x):=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(|x|<1)\ .$ Using termwise differentiation one obtains $f'(x)=\sum_{k=1}^\infty {\alpha\choose k}\,k\, x^{k-1}=\sum_{k'=0}^\infty {\alpha\choose k'+1}\,(k'+1)\, x^{k'}=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k\ .$ Therefore, using the third and the first expression for $f'(x)$, we get $(1+x)f'(x)=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k+\sum_{k=0}^\infty {\alpha\choose k}\,k\, x^k=\alpha f(x)\ ,$ or $(1+x)f'(x)-\alpha f(x)\equiv0\ .$ It follows that ${d\over dx}\bigl((1+x)^{-\alpha} f(x)\bigr)\equiv0$, or $(1+x)^{-\alpha} f(x)={\rm const.}$, and as $f(0)=1$ one concludes that $f(x)\equiv (1+x)^\alpha$; whence $(1)$ is indeed true for $|x|<1$.

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    @Milosz: See my edit.2012-04-27
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In addition (and for future reference) to everybody else's answers I have found the following to be true:

$ \begin{align} (x+y)^n&=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(1)\\ &=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(2) \end{align} $

  • Let $x,y\in\mathbb{C}\setminus\{0\}$. Eq. $(1)$ and $(2)$ are both true if and only if $n\in\mathbb N$.
  • For $n\in\mathbb C$ and for $(2)$ to hold, the values of $x$ and $y$ must satisfy $|x|>|y|$.

A quick look at the case $y=1$ tells us that: $ \begin{align} (1+x)^n&=\sum^{\infty}_{r=0}\binom{n}{r}x^r &\qquad(3)\\ &=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r} &\qquad(4) \end{align} $

  • When $n\in\mathbb{C}$ and |x|<1 we'd use $(3)$
  • When $n\in\mathbb{C}$ and $|x|>1$ we'd use $(4)$
  • When $n\in\mathbb{N}$ or $|x|=1$ either equation works
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In case you don't know the Taylor-expansion yet:
1) For $n \in \mathbb{N}$ and n < m we have $\binom{n}{m}$ = 0. Therefore further summation does not change the result.
2) For $n$ a fraction like -1/2 (and y = 1, |x| < 1) you can use the sum of the geometric series to get an infinite series $ \sqrt{\dfrac{1}{1+x}}\sqrt{\dfrac{1}{1+x}} = 1 - 1x^1 + 1x^2 +- ... = $ To solve this, put $ \sqrt{\dfrac{1}{1+x}}= a_0 + a_1x^1 + a_2x^2 + ... $ and multiply this with itself, ordering the result by exponents of $x$ and comparing with the coefficients 1 and -1, respectively, of the upper series.

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    Of course you are right. Thanks2012-04-24