题目 设函数 $f_n(x)=x^n(1-x)^2$ 在 $\left[\dfrac12,1\right]$ 上的最大值为 $a_n(n=1,2,3,\cdots)$． 问题1 求数列 $\{a_n\}$ 的通项公式； 答案1 $a_n=\begin{cases}\dfrac18,&n=1,\\ \dfrac{4n^n}{(n+2)^{n+2}},&n\geqslant2.\end{cases}$ 解析1 对原函数求导，得$$\begin{split}f'_n(x)&=nx^{n-1}(1-x)^2-2x^n(1-x)\\ &=x^{n-1}(1-x)[n(1-x)-2x],\end{split}$$由 $f'_n(x)=0$ 知 $x=1$ 或者 $x=\dfrac{n}{n+2}$，又 $f_n(1)=0$．<br/><case>情形一</case> 当 $n=1$ 时，$$\dfrac{n}{n+2}=\dfrac13\notin\left[\dfrac12,1\right],$$又$$f_1\left(\dfrac12\right)=\dfrac18>0,$$故 $a_1=\dfrac18$；<br/><case>情形二</case> 当 $n=2$ 时，$$\dfrac{n}{n+2}=\dfrac12\in\left[\dfrac12,1\right],$$又$$f_2\left(\dfrac12\right)=\dfrac{1}{16}>0,$$故 $a_2=\dfrac{1}{16}$；<br/><case>情形三</case> 当 $n\geqslant3$ 时，$$\dfrac{n}{n+2}\in\left[\dfrac12,1\right],$$因为 $x\in\left[\dfrac12,\dfrac{n}{n+2}\right)$ 时，$f'_n(x)>0$；$x\in\left(\dfrac{n}{n+2},1\right)$ 时，$f'_n(x)<0$．<br/>所以 $f_n(x)$ 在 $x=\dfrac{n}{n+2}$ 处取得最大值，即$$a_n=\left(\dfrac{n}{n+2}\right)^n\left(\dfrac{2}{n+2}\right)^2=\dfrac{4n^n}{(n+2)^{n+2}},$$检验知，$a_2=\dfrac {1}{16}$ 也满足此式．<br/>综上所述，$$a_n=\begin{cases}\dfrac18,&n=1,\\ \dfrac{4n^n}{(n+2)^{n+2}},&n\geqslant2.\end{cases}$$ 备注1 问题2 求证：对任何正整数 $n(n\geqslant2)$，都有 $a_n\leqslant\dfrac{1}{(n+2)^2}$ 成立； 答案2 略 解析2 当 $n\geqslant2$ 时，欲证$$\dfrac{4n^n}{(n+2)^{n+2}}\leqslant\dfrac{1}{(n+2)^2},$$只需证明 $\left(1+\dfrac2n\right)^n\geqslant4$．<br/>因为$$\begin{split}\left(1+\dfrac2n\right)^n&=\mathrm{C}_n^0+\mathrm{C}_n^1\cdot\left(\dfrac{2}{n}\right)^1+\cdots+\mathrm{C}_n^n\cdot\left(\dfrac{2}{n}\right)^n\\&\geqslant1+2+\dfrac{n(n-1)}{2}\cdot\dfrac{4}{n^2}\\&\geqslant1+2+1=4,\end{split}$$所以当 $n\geqslant2$ 时，都有 $a_n\leqslant \dfrac{1}{(n+2)^2}$ 成立． 备注2 问题3 设数列 $\{a_n\}$ 的前 $n$ 项和为 $S_n$，求证：对任意正整数 $n$，都有 $S_n<\dfrac{7}{16}$ 成立． 答案3 略 解析3 当 $n=1,2$ 时，结论显然成立；<br/>当 $n\geqslant3$ 时，由 $(2)$ 知\[\begin{split}S_n&=\dfrac18+\dfrac{1}{16}+a_3+a_4+\cdots+a_n\\&<\dfrac18+\dfrac{1}{16}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\cdots+\dfrac{1}{(n+2)^2}\\&<\dfrac18+\dfrac{1}{16}+\left(\dfrac14-\dfrac15\right)+\left(\dfrac15-\dfrac16\right)+\cdots+\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)\\&<\dfrac18+\dfrac{1}{16}+\dfrac14\\&=\dfrac{7}{16}.\end{split}\]因此，对任意正整数 $n$，都有 $S_n<\dfrac{7}{16}$ 成立． 备注3