证明恒等式：$\displaystyle \sum\limits_{i=0}^n\dbinom{n}{i}\dbinom{n+i}{i}=\sum\limits_{i=0}^n2^i\dbinom{n}{i}^2$
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【答案】
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【解析】
\begin{align}\sum\limits_{i=0}^{n}\dbinom{n}{i}\dbinom{n+i}{n}&=\sum_{i=0}^n\left[\dbinom{n}{i}\sum_{k=0}^n\dbinom{n}{k}\dbinom{i}{n-k}\right]\\
&=\sum_{i=0}^n\left[\dbinom{n}{k}\sum{i=0}^n\dbinom{n}{i}\dbinom{i}{n-k}\right]\\
&=\sum_{k=0}^n\left[\dbinom{n}{k}\sum_{i=n-k}^n\dbinom{n}{i}\dbinom{i}{n-k}\right]\\
&=\sum_{k=0}^n\dbinom{n}{k}2^k\dbinom{n}{n-k}\\
&=\sum_{k=0}^n2^k\dbinom{n}{k}^2=\sum_{i=0}^n2^i\dbinom{n}{i}^2
\end{align}
&=\sum_{i=0}^n\left[\dbinom{n}{k}\sum{i=0}^n\dbinom{n}{i}\dbinom{i}{n-k}\right]\\
&=\sum_{k=0}^n\left[\dbinom{n}{k}\sum_{i=n-k}^n\dbinom{n}{i}\dbinom{i}{n-k}\right]\\
&=\sum_{k=0}^n\dbinom{n}{k}2^k\dbinom{n}{n-k}\\
&=\sum_{k=0}^n2^k\dbinom{n}{k}^2=\sum_{i=0}^n2^i\dbinom{n}{i}^2
\end{align}
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