已知数列 $\{a_n\}$ 满足 $a_n=\frac{(n^2+1)^2}{n^4+4}$（$n=1,2,3,\ldots$）．证明：对任意正整数 $n$，都有 $a_1^2a_2^{n-1}a_3^{n-2}\ldots a_n=\frac{2^{n+1}}{n^2+2n+2}$．
【难度】
【出处】
全国高中数学联赛模拟试题(10)
【标注】
【答案】
略
【解析】
设 $c_k=k^2+1,d_k=k^4+1$，则$$c_{k-1}c_{k+1}=((k-1)^2+1)((k+1)^2+1)=(k^2-2k+2)(k^2+2k+2)=k^4+4=d_k,$$即$$a_k=\frac{c_k^2}{c_{k-1}c_{k+1}}.$$从而，对任意正整数 $n$，有$$\begin{aligned}
a_1^na_2^{n-1}a_3^{n-2}\ldots a_n&=\prod^n_{k=1}a_k^{n+1-k}=\prod^n_{k=1}\frac{c_k^{2n+2-2k}}{c_{k-1}^{n+1-k}c_{k+1}^{n+1-k}}\\
&=\frac{c_1^{2n-(n-1)}}{c_0^n}\cdot \prod^n_{k=2}\frac{c_k^{2n+2-2k}}{c_k^{n+2-k}c_k^{n-k}}\cdot \frac{1}{c_{n+1}}\\
&=\frac{c_1^{n+1}}{c_0^n}\cdot 1^{n-1}\cdot \frac{1}{c_{n+1}}\\
&=\frac{c_1^{n+1}}{c_0^nc_{n+1}}.\\
\end{aligned}$$由 $c_0=1, c_1=2, c_{n+1}=(n+1)^2+1=n^2+2n+2$，得$$a_1^na_2^{n-1}a_3^{n-2}\ldots a_n=\frac{2^{n+1}}{n^2+2n+2}.$$
a_1^na_2^{n-1}a_3^{n-2}\ldots a_n&=\prod^n_{k=1}a_k^{n+1-k}=\prod^n_{k=1}\frac{c_k^{2n+2-2k}}{c_{k-1}^{n+1-k}c_{k+1}^{n+1-k}}\\
&=\frac{c_1^{2n-(n-1)}}{c_0^n}\cdot \prod^n_{k=2}\frac{c_k^{2n+2-2k}}{c_k^{n+2-k}c_k^{n-k}}\cdot \frac{1}{c_{n+1}}\\
&=\frac{c_1^{n+1}}{c_0^n}\cdot 1^{n-1}\cdot \frac{1}{c_{n+1}}\\
&=\frac{c_1^{n+1}}{c_0^nc_{n+1}}.\\
\end{aligned}$$由 $c_0=1, c_1=2, c_{n+1}=(n+1)^2+1=n^2+2n+2$，得$$a_1^na_2^{n-1}a_3^{n-2}\ldots a_n=\frac{2^{n+1}}{n^2+2n+2}.$$
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