题目 已知 $f(x)$ 为二次函数，不等式 $f(x)+2<0$ 的解集为 $\left(-1,\dfrac{1}{3}\right)$，且对任意 $\alpha,\beta\in\mathbb R$，恒有 $f(\sin\alpha)\leqslant 0$，$f(2+\cos \beta)\geqslant 0$．数列 $\{a_{n}\}$ 满足 $a_{1}=1$，$3a_{n+1}=1-\dfrac{1}{f'(a_{n})}(n\in\mathbb N^{*})$． 问题1 求函数 $f(x)$ 的解析式； 答案1 $f(x)=\dfrac{3}{2}x^{2}+x-\dfrac{5}{2}$ 解析1 依题意，$f(x)+2=a(x+1)\left(x-\dfrac{1}{3}\right)(a>0)$，即\[f(x)=ax^{2}+\dfrac{2a}{3}x-\dfrac{a}{3}-2.\]令 $\alpha=\dfrac{\pi}{2}$，$\beta=\pi$，则$$\sin\alpha=1 , \cos\beta=-1,$$由$$f(1)\leqslant 1 , f(2-1)\geqslant 0,$$得\[f(1)=0,\]即$$a+\dfrac{2a}{3}-\dfrac{a}{3}-2=0,$$解得 $a=\dfrac{3}{2}$，因此$$f(x)=\dfrac{3}{2}x^{2}+x-\dfrac{5}{2}.$$ 备注1 问题2 设 $b_{n}=\dfrac{1}{a_{n}}$，求数列 $\{b_{n}\}$ 的通项公式； 答案2 $b_{n}=3n-2,n\in\mathbb N^{*}$ 解析2 $f'(x)=3x+1$，则\[\begin{split}3a_{n+1}&=1-\dfrac{1}{f'(a_{n})}\\ &=1-\dfrac{1}{3a_{n}+1}\\ &=\dfrac{3a_{n}}{3a_{n}+1},\end{split}\]即$$a_{n+1}=\dfrac{a_{n}}{3a_{n}+1},$$两边取倒数，得\[\dfrac{1}{a_{n+1}}=3+\dfrac{1}{a_{n}},\]即 $b_{n+1}=3+b_{n}$．<br/>所以数列 $\{b_{n}\}$ 是首项为 $b_{1}=\dfrac{1}{a_{1}}=1$，公差为 $3$ 的等差数列，所以$$b_{n}=1+(n-1)\cdot 3=3n-2(n\in\mathbb N^{*}).$$ 备注2 问题3 若 $(2)$ 中数列 $\{b_{n}\}$ 的前 $n$ 项和为 $S_{n}$，求数列 $\left\{S_{n}\cdot \cos\left(b_{n}\pi\right)\right\}$ 的前 $n$ 项和 $T_{n}$． 答案3 $T_{n}=\begin{cases}\dfrac{-3n^{2}-2n+1}{4},&2\nmid n,\\ \dfrac{3n^{2}+2n}{4},&2\mid n.\end{cases}$ 解析3 因为$$\cos(b_{n}\pi)=\cos(3n-2)\pi=\cos(n\pi)=(-1)^{n},$$所以$$S_{n}\cdot \cos(b_{n}\pi)=(-1)^{n}\cdot S_{n},$$故\[T_{n}=-S_{1}+S_{2}-S_{3}+S_{4}-\cdots+(-1)^{n}S_{n}.\]<case>情形一</case> 当 $n$ 为偶数时，\[\begin{split}T_{n}&=(S_{2}-S_{1})+(S_{4}-S_{3})+\cdots+(S_{n}-S_{n-1})\\&=b_{2}+b_{4}+\cdots+b_{n}\\&=\dfrac{\dfrac{n}{2}(b_{2}+b_{4})}{2}\\&=\dfrac{n}{4}(4+3n-2)\\&=\dfrac{3n^{2}+2n}{4}.\end{split}\]<case>情形二</case> 当 $n$ 奇数时，\[\begin{split}T_{n}&=T_{n-1}-S_{n}\\ &=\dfrac{3(n-1)^{2}+2(n-1)}{4}-\dfrac{n(1+3n-2)}{2}\\ &=\dfrac{-3n^{2}-2n+1}{4}.\end{split}\]综上，$T_{n}=\begin{cases}\dfrac{-3n^{2}-2n+1}{4},&2\nmid n,\\ \dfrac{3n^{2}+2n}{4},&2\mid n.\end{cases}$ 备注3