I have been trying to understand why the binomial theorem can work for negative and fractional indices.
I understand that when raising binomials to positive integral indices, each coefficient is simply the number of ways that you can pick each term (e.g. for $(x+y)^5$, if you want to make up $xy^4$ there are 5 brackets from which to pick the $x$, so this term will come up 5 times in the full expansion).
I am not sure if a way to understand the infinite expansion for negative and fractional indices exists, but if it does I would very much like to know! Otherwise, I haven't been able to find a proof that shows that the result of the expansion for positive integer powers is valid for negative or fractional indices. I would be very grateful if someone could point me in the direction of such a proof.
Best Answer
This is probably the wrong proof for you, but I will post it anyways. (requires calculus)
Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.
If it is an analytic function, then it should follow Taylor's theorem.
Now, if we take the expansion around $x=0$, we get
$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$
Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$
or
$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$
$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$
where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.