begin{split} (1+x)^{alpha}=&sum_{n=0}^{infty}frac{alpha(alpha-1)(alpha-2)cdots(alpha-n+1)}{n!}x^n/ =&1+alpha x+frac{alpha(alpha-1)}{2!}x^2+cdots+frac{alpha(alpha-1)(alpha-2)cdots(alpha-n+1)}{n!}x^n+cdots.~~~~~~~(1) end{split} /

证明：

• 首先确定右端幂级数的收敛半径R.
• 由于

rho=lim_{nrightarrowinfty}|frac{a_{n+1}}{a_n}|=lim_{nrightarrowinfty}frac{|alpha-n|}{n+1}=1, /

所以这个幂级数的收敛半径为R=frac{1}{rho}=1.

• 用S_{alpha}(x)表示右端幂级数的和函数

S_{alpha}(x)=sum_{n=0}^{infty}frac{alpha(alpha-1)(alpha-2)cdots(alpha-n+1)}{n!}x^n,~-1<x<1./

• 可知S_{alpha}(0)=1.
• 逐项求导，得

S'_{alpha}(x)=sum_{n=1}^{infty}frac{alpha(alpha-1)(alpha-2)cdots(alpha-n+1)}{(n-1)!}x^{n-1}=alpha S_{alpha-1}(x),~-1<x<1./

• 因此有

(1+x)S'_{alpha}(x)=alpha(1+x)S_{alpha-1}(x),~-1<x<1./

• 而

begin{split} (1+x)S_{alpha-1}(x)=&(1+x)sum_{n=0}^{infty}frac{(alpha-1)(alpha-2)cdots(alpha-n)}{n!}x^n/ =&sum_{n=0}^{infty}frac{(alpha-1)(alpha-2)cdots(alpha-n)}{n!}x^n+sum_{n=0}^{infty}frac{(alpha-1)(alpha-2)cdots(alpha-n)}{n!}x^{n+1}/ =&1+sum_{n=1}^{infty}[frac{(alpha-1)(alpha-2)cdots(alpha-n)}{n!}+frac{(alpha-1)(alpha-2)cdots(alpha-n+1)}{(n-1)!}]x^n/ =&1+sum_{n=1}^{infty}frac{(alpha-1)(alpha-2)cdots(alpha-n+1)}{(n-1)!}[frac{alpha-n}{n}+1]x^n/ =&sum_{n=0}^{infty}frac{alpha(alpha-1)(alpha-2)cdots(alpha-n+1)}{n!}x^n=S_{alpha}(x),~-1<x<1. end{split}

• 因此

(1+x)S'_{alpha}(x)=alpha S_{alpha}(x), /

也即

S'_{alpha}(x)=frac{alpha}{1+x} S_{alpha}(x),~-1<x<1./

• 解常微分方程（见林伟初高数下册教材p.120），得

S_{alpha}=Ce^{int_0^xfrac{alpha}{1+t}dt}=Ce^{alphaln(1+x)}=C(1+x)^{alpha},-1<x<1./

• 由于S_{alpha}(0)=1，代入上式求得C=1，所以

S_{alpha}(x)=(1+x)^{alpha},-1<x<1/

命题得证.

注记：

• 等式(1)叫做牛顿二项公式，右端的幂级数叫做牛顿二项级数.
• 当alpha为正整数时，就是通常大家高中知道的牛顿二项（式展开）公式，比如alpha=m，此时

(1+x)^m=sum_{n=0}^{m}C_m^nx^n, /

其中

C_m^n=frac{m!}{n!(m-n)!}=frac{m(m-1)cdots(m-n+1)}{n!},0leq nleq m, / C_m^n=0,ngeq m+1. /

• 在端点x=pm1时等式是否成立，比较复杂，依赖于alpha的具体取值.