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由题意, 连续函数 \(f(x)=\begin{cases}a_n x^n,& \frac{1}{n+1}< x\le\frac{1}{n};\\ a_{n+1}x^{n+1},& \frac{1}{n+2}< x\le\frac{1}{n+1}.\end{cases}\)
\(\therefore\quad a_{n+1}(\frac{1}{n+1})^{n+1}=f(\frac{1}{n+1})=f(\frac{1}{n+1}{\small+})=a_n(\frac{1}{n+1})^n\)
\(\quad\frac{a_{n+1}}{n+1}=a_n,\;\frac{a_{n+1}}{(n+1)!}=a_1=1,\;\boxed {a_n = n!\small\,(n=1,2,\ldots)}\) |
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