问题及解答

证明: 多项式函数 $f(x)=(1+x)^{2n+1}$ 在 $x=0$ 处可展开为

\[
(1+x)^{2n+1}=\sum_{k=0}^{n}C_{2n+1}^k x^k+R_n(x).
\]

首先计算 $f$ 的各阶导数.\pause
\[
\begin{aligned}
f'(x)&=(2n+1)(1+x)^{2n},\\
f''(x)&=(2n+1)(2n)(1+x)^{2n-1},\\
f'''(x)&=(2n+1)(2n)(2n-1)(1+x)^{2n-2},\\
&\vdots\\
f^{(n)}(x)&=(2n+1)(2n)(2n-1)\cdots(n+1)(1+x)^n.
\end{aligned}
\]

于是
\[
\begin{aligned}
f'(0)&=2n+1,\\
f''(0)&=(2n+1)(2n),\\
f'''(0)&=(2n+1)(2n)(2n-1),\\
&\vdots\\
f^{(n)}(0)&=(2n+1)(2n)(2n-1)\cdots(n+1).
\end{aligned}
\]

因此,
\[
\begin{split}
(1+x)^{2n+1}&=f(0)+\frac{f'(0)}{1!}x^1+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+R_n(x)\\
&=1+\frac{2n+1}{1}x^1+\frac{(2n+1)(2n)}{2!}x^2+\cdots\\
&\qquad +\frac{(2n+1)(2n)(2n-1)\cdots(n+1)}{n!}x^n+R_n(x)\\
&=1+C_{2n+1}^1 x+C_{2n+1}^2 x^2+\cdots+C_{2n+1}^n x^n+R_n(x)\\
&=\sum_{k=0}^{n}C_{2n+1}^k x^k+R_n(x).
\end{split}
\]