已知数列 $\left\{ {a_n} \right\}$ 与 $\left\{ {b_n} \right\}$ 满足：${b_n}{a_n} + {a_{n + 1}} + {b_{n + 1}}{a_{n + 2}} = 0$，${b_n} = \dfrac{{3 + {{\left( - 1\right)}^n}}}{2}$，$n \in {{\mathbb{N}}^ * }$，且 ${a_1} = 2$，${a_2} = 4$．
【难度】
【出处】
无
【标注】
-
求 ${a_3}$，${a_4}$，${a_5}$ 的值；标注答案$a_3=-3$，$a_4=-5$，$a_5=4$解析当 $n$ 为奇数时，$b_n=1$；当 $n$ 为偶数时，$b_n=2$．
于是 $a_3=-3$，$a_4=-5$，$a_5=4$． -
设 ${c_n} = {a_{2n - 1}} + {a_{2n + 1}}$，$n \in {{\mathbb{N}}^ * }$，证明：$\left\{ {c_n} \right\}$ 是等比数列；标注答案略解析对任意 $ n \in {\mathbb{N}}^* $，\[\begin{split}a_{2n - 1} + a_{2n} + 2a_{2n + 1} = 0, &\quad \cdots \cdots {\text{ ① }} \\
2a_{2n} + a_{2n + 1} + a_{2n + 2} = 0, &\quad \cdots \cdots {\text{ ② }} \\
a_{2n + 1} + a_{2n + 2} + 2a_{2n + 3} = 0,&\quad \cdots \cdots {\text{ ③ }}\end{split}\]② $- $ ③，得\[ a_{2n} = a_{2n + 3} . \quad \cdots \cdots {\text{ ④ }} \]将 ④ 代入 ①，可得\[ a_{2n + 1} + a_{2n + 3} = - \left(a_{2n - 1} + a_{2n + 1} \right),\]即\[ c_{n + 1} = - c_n \left(n \in {\mathbb{N}}^* \right) .\]又 $ c_1 = a_1 + a_3 = - 1 $，故 $ c_n \ne 0 $，因此\[ \frac{{c_{n + 1} }}{c_n } = - 1, \]所以 $ \left\{ c_n \right\} $ 是等比数列． -
设 ${S_k} = {a_2} + {a_4} + \cdots + {a_{2k}}$，$k \in {{\mathbb{N}}^ * }$，证明：$\displaystyle \sum\limits_{k = 1}^{4n} {\dfrac{S_k}{a_k} < \dfrac{7}{6}\left(n \in {{\mathbb{N}}^ * }\right)} $．标注答案略解析由（2）可得\[ a_{2k - 1} + a_{2k + 1} = \left( - 1\right)^k , \]于是，对任意 $ k \in{\mathbb{ N}}^* $ 且 $ k \geqslant 2 $，有\[\begin{split}
a_1 + a_3 = - 1, \\
- \left(a_3 + a_5 \right) = - 1, \\
a_5 + a_7 = - 1, \\
\vdots \\
\left( - 1\right)^k \left(a_{2k - 3} + a_{2k - 1} \right) = - 1. \\
\end{split} \]将以上各式相加，得\[ a_1 + \left( - 1\right)^k a_{2k - 1} = - \left(k - 1\right), \]即\[ a_{2k - 1} = \left( - 1\right)^{k + 1} \left(k + 1\right), \]此式当 $ k=1 $ 时也成立．由 ④ 式得\[ a_{2k} = \left( - 1\right)^{k + 1} \left(k + 3\right). \]从而\[ \begin{split}S_{2k} & = \left(a_2 + a_4 \right) + \left(a_6 + a_8 \right) + \cdots + \left(a_{4k - 2} + a_{4k} \right) = - k,\\
S_{2k - 1} &= S_{2k} - a_{4k} = k + 3.\end{split} \]所以，对任意 $ n \in {\mathbb{N}}^* $，$ n \geqslant 2 $，\[\begin{split}\sum\limits_{k = 1}^{4n} \frac{S_k }{a_k } & = \sum\limits_{m = 1}^n {\left(\frac{{S_{4m - 3} }}{{a_{4m - 3} }} + \frac{{S_{4m - 2} }}{{a_{4m - 2} }} + \frac{{S_{4m - 1} }}{{a_{4m - 1} }} + \frac{{S_{4m} }}
{{a_{4m} }}\right)} \\&
= \sum\limits_{m = 1}^n {\left(\frac{2m + 2}{2m} - \frac{2m - 1}{2m + 2} - \frac{2m + 3}{2m + 1} + \frac{2m}{2m + 3}\right)} \\&
= \sum\limits_{m = 1}^n {\left(\frac{2}{2m\left(2m + 1\right)} + \frac{3}{\left(2m + 2\right)\left(2m + 3\right)}\right)} .\end{split}\]而当 $n\geqslant 3$ 时，\[\begin{split}& \sum\limits_{m=1}^{n}{\dfrac 2{2m(2m+1)}}<\dfrac 13 + \sum\limits_{m=2}^{n}{\dfrac {2}{(2m-1)(2m+1)}}=\dfrac 13 + \sum\limits_{m=2}^{n}\left(\dfrac 1{2m-1}-\dfrac 1{2m+1}\right)=\dfrac 23 -\dfrac 1{2n+1},\\& \sum\limits_{m=1}^{n}{\dfrac 3{(2m+2)(2m+3)}}< \sum\limits_{m=1}^{n}{\dfrac 3{(2m+1)(2m+3)}}=\dfrac 32 \cdot \sum\limits_{m=1}^{n}\left(\dfrac 1{2m+1}-\dfrac 1{2m+3}\right)=\dfrac 12 -\dfrac 32\cdot \dfrac 1{2n+3}.\end{split}\]所以原式 $<\dfrac 23 -\dfrac 1{2n+1}+\dfrac 12 -\dfrac 32 \cdot \dfrac 1{2n+3}<\dfrac 76$．
经检验，当 $n=1,2$ 时，不等式均成立．
综上，原不等式得证．
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