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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:$\displaystyle \sum_{k=1}^{4n}\dfrac{S_k}{a_k}<\dfrac 76$($n\in\mathcal N^*$).


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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.

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