问题及解答

证明恒等式

\[
\dfrac{C_n^0}{2^n}+\dfrac{C_{n+1}^0}{2^{n+1}}+\dfrac{C_{n+2}^0}{2^{n+2}}+\cdots+\dfrac{C_{2n}^0}{2^{2n}}=1.
\]

证明: (使用归纳法) 当 $n=1,2$ 时, 等式显然成立. 假设所证恒等式对于 $n=k$ 时成立, 即

\[
\dfrac{C_k^0}{2^k}+\dfrac{C_{k+1}^0}{2^{k+1}}+\dfrac{C_{k+2}^0}{2^{k+2}}+\dfrac{C_{k+3}^0}{2^{k+3}}+\cdots+\dfrac{C_{k+k}^0}{2^{k+k}}=1.
\]

当 $n=k+1$ 时, 记

\[
A=\dfrac{C_{k+1}^0}{2^{k+1}}+\dfrac{C_{k+2}^1}{2^{k+2}}+\dfrac{C_{k+3}^2}{2^{k+3}}+\cdots+\dfrac{C_{2k+1}^k}{2^{2k+1}}+\dfrac{C_{2k+2}^{k+1}}{2^{2k+2}},
\]

利用恒等式 $C_{n+1}^{m}=C_n^m+C_n^{m-1}$, 可得

\[
\begin{split}
A&=\frac{1}{2^{k+1}}+\frac{C_{k+1}^1+C_{k+1}^0}{2\cdot 2^{k+1}}+\frac{C_{k+2}^2+C_{k+2}^1}{2\cdot 2^{k+2}}+\frac{C_{k+3}^3+C_{k+3}^2}{2\cdot 2^{k+3}}+\cdots+\frac{C_{2k}^k+C_{2k}^{k-1}}{2\cdot 2^{2k}}+\frac{C_{2k+1}^{k+1}+C_{2k+1}^k}{2\cdot 2^{2k+1}}\\
&=\frac{1}{2}\cdot\biggl[\frac{1}{2^k}+\frac{C_{k+1}^1}{2^{k+1}}+\frac{C_{k+2}^2}{2^{k+2}}+\frac{C_{k+3}^3}{2^{k+3}}+\cdots+\frac{C_{2k}^k}{2^{2k}}\biggr]+\frac{1}{2}\biggl[\frac{C_{k+1}^0}{2^{k+1}}+\frac{C_{k+2}^1}{2^{k+2}}+\frac{C_{k+3}^2}{2^{k+3}}+\cdots+\frac{C_{2k}^{k-1}}{2^{2k}}+\frac{C_{2k+1}^k}{2^{2k+1}}\biggr]+\frac{C_{2k+1}^{k+1}}{2\cdot 2^{2k+1}}\\
&=\frac{1}{2}+\frac{1}{2}\biggl[A-\frac{C_{2k+2}^{k+1}}{2^{2k+2}}\biggr]+\frac{C_{2k+1}^{k+1}}{2\cdot 2^{2k+1}}\\
&=\frac{1}{2}+\frac{1}{2}A+\frac{1}{2}\biggl[\frac{C_{2k+1}^{k+1}}{2^{2k+1}}-\frac{C_{2k+2}^{k+1}}{2^{2k+2}}\biggr]\\
&=\frac{1}{2}+\frac{1}{2}A,
\end{split}
\]

由此得 $A=1$. 故恒等式对 $n=k+1$ 也成立. 证毕.