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:

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

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

\[Q.E.D.\]