2011年高考天津卷理科数学第20题（压轴题）：
设数列$\{a_n\}$和$\{b_n\}$满足：$b_na_n+a_{n+1}+b_{n+1}a_{n+2}=0$，$b_n=\dfrac{3+(-1)^n}2$，$n\in\mathcal N^*$，且$a_1=2$，$a_2=4$．
(1) 求$a_3,a_4,a_5$的值；
(2) 设$c_n=a_{2n-1}+a_{2n+1}$，$n\in\mathcal N^*$，求证：$\{c_n\}$是等比数列；
(3) 设$S_k=a_2+a_4+\cdots +a_{2k}$，$k\in\mathcal N^*$，求证：$\displaystyle \sum_{k=1}^{4n}\dfrac{S_k}{a_k}<\dfrac 76$($n\in\mathcal N^*$)．
分析与解 (1) 当$n$为奇数时，$b_n=1$；当$n$为偶数时，$b_n=2$，于是$a_3=-3$，$a_4=-5$，$a_5=4$．
(2) 对任意$n\in\mathcal N^*$，有\[\begin{split} a_{2n-1}+a_{2n}+2a_{2n+1}&=0,\\2a_{2n}+a_{2n+1}+a_{2n+2}&=0,\\a_{2n+1}+a_{2n+2}+2a_{2n+3}&=0,\end{split} \]
从而可得$a_{2n}=a_{2n+3}$，于是$$\dfrac{c_{n+1}}{c_n}=\dfrac{a_{2n+1}+a_{2n+3}}{a_{2n-1}+a_{2n+1}}=\dfrac{a_{2n}+a_{2n+1}}{a_{2n-1}+a_{2n+1}}=-1,$$所以$\{c_n\}$是等比数列．
(3) 根据(2)，$a_{2k-1}+a_{2k+1}=(-1)^k$，$k\in\mathcal N^*$，于是不难求得$$a_n=\begin{cases} (-1)^{k+1}(k+1),&n=2k-1,\\ (-1)^{k+1}(k+3),&n=2k,\end{cases} $$从而$$S_{2k}=-k,S_{2k-1}=k+3,$$于是\[\begin{split} \sum_{k=1}^{4n}\dfrac{S_k}{a_k}&=\sum_{m=1}^n\left(\dfrac{S_{4m-3}}{a_{4m-3}}+\dfrac{S_{4m-2}}{a_{4m-2}}+\dfrac{S_{4m-1}}{a_{4m-1}}+\dfrac{S_{4m}}{a_{4m}}\right)\\&=\sum_{m=1}^n\left(\dfrac{2m+2}{2m}-\dfrac{2m-1}{2m+2}-\dfrac{2m+3}{2m+1}+\dfrac{2m}{2m+3}\right)\\&=\sum_{m=1}^n\left[\dfrac{2}{2m(2m+1)}+\dfrac{3}{(2m+2)(2m+3)}\right],\end{split} \]而当$n\geqslant 3$时，有$$\begin{split} \sum_{m=1}^n\dfrac{2}{2m(2m+1)}<&\dfrac 13+\sum_{m=2}^n\dfrac{2}{(2m-1)(2m+1)}\\=&\dfrac 13+\sum_{m=2}^n\left(\dfrac 1{2m-1}-\dfrac{1}{2m+1}\right)\\=&\dfrac 23-\dfrac 1{2n+1},\end{split} $$又$$\begin{split} \sum_{m=1}^n\dfrac{3}{(2m+2)(2m+3)}<&\sum_{m=1}^n\dfrac{3}{(2m+1)(2m+3)}\\=&\dfrac 32\cdot \sum_{m=1}^n\left(\dfrac 1{2m+1}-\dfrac{1}{2m+3}\right)\\=&\dfrac 12-\dfrac 32\cdot \dfrac 1{2n+3},\end{split} $$于是$$\sum_{k=1}^{4n}\dfrac{S_k}{a_k}<\dfrac 23-\dfrac{1}{2n+1}+\dfrac 12-\dfrac 32\cdot \dfrac{1}{2n+3}<\dfrac 76,$$经检验当$n=1,2$时不等式均成立．
综上所述，原不等式得证．