## Proof: Binomial Expansion for Negative and Fractional n Formula

For negative and fractional n, the binomial expansion of $$(1+x)^n$$ is given by:

$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots +\frac{n(n-1) \cdots (n-r+1)}{r!}x^r+ \cdots=\sum_{r=0}^{\infty} \frac{n(n-1) \cdots (n-r+1)}{r!}x^r$

Given that abs(x)<1 .

## Proof

We can derive this from the Maclaurin series of the function. The Maclaurin series of a function, $$f(x)$$, is given by:

$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!} x^2 +\frac{f'''(0)}{3!} x^3 + \cdots + \frac{f^{(r)}(0)}{r!} x^r +\cdots=\sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} x^r$

If we let $$f(x)=(1+x)^n$$, then:

$f'(x)=n(1+x)^{n-1}$

$f''(x)=n(n-1)(1+x)^{n-2}$

$\vdots$

$f^{(r)}(x)=n(n-1) \cdots (n-r+1)(1+x)^{n-r}$

Therefore the Maclaurin series for $$(1+x)^n$$ is:

$(1+x)^n=\sum_{r=0}^{\infty} \frac{n(n-1) \cdots (n-r+1)(1+0)^{n-r}}{r!} x^r$

$(1+x)^n=\sum_{r=0}^{\infty} \frac{n(n-1) \cdots (n-r+1)}{r!} x^r$

$Q.E.D.$