在数列 $\{a_{n}\}$ 中，$a_{1}=0$，且对任意 $k\in\mathbb N^{*}$，$a_{2k-1},a_{2k},a_{2k+1}$ 成等差数列，其公差为 $d_{k}$．
【难度】
【出处】
2010年高考天津卷（理）
【标注】
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若 $d_{k}=2k$，证明 $a_{2k},a_{2k+1},a_{2k+2}$ 成等比数列（$k\in\mathbb N^{*}$）；标注答案略解析由题设 ${a_{2k + 1}} - {a_{2k - 1}} = 4k,k \in {{\mathbb{N}}^*}$．于是累加可得\[ {a_{2k + 1}} - {a_1} =2k\left(k+1\right),\]所以 $a_{2k+1}=2k(k+1)$，因此\[ \dfrac{{{a_{2k + 1}}}}{{{a_{2k}}}} =\dfrac{2k(k+1)}{2k(k+1)-2k}= \dfrac{k + 1}{k},\dfrac{{{a_{2k + 2}}}}{{{a_{2k + 1}}}} =\dfrac{2k(k+1)+2(k+1)}{2k(k+1}= \dfrac{k + 1}{k},\]所以 $a_{2k},a_{2k+1},a_{2k+2}$ 成等比数列．
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若对任意 $k \in {{\mathbb{N}}^*}$，${a_{2k}}$，${a_{2k + 1}}$，${a_{2k + 2}}$ 成等比数列，其公比为 ${q_k}$．
（i）设 ${q_1} \ne 1$．证明 $\left\{ {\dfrac{1}{{{q_k} - 1}}} \right\}$ 是等差数列；
（ii）若 ${a_2} = 2$，证明 $\displaystyle \dfrac{3}{2} < 2n - \sum\limits_{k = 2}^n {\dfrac{k^2}{a_k}} \leqslant 2\left(n \geqslant 2\right)$．标注答案略解析由题设 $\dfrac{a_{2k+2}}{a_{2k}}=q_{k}^{2}$，于是$$a_{2k}=q_{k-1}^{2}\cdot q_{k-2}^{2}\cdots q_{1}^{2}\cdot a_{2},$$所以\[a_{2k+1}=q_{k}\cdot q_{k-1}^{2}\cdot q_{k-2}^{2}\cdots q_{1}^{2}\cdot a_{2},a_{2k-1}=q_{k-1}\cdot q_{k-2}^{2}\cdots q_{1}^{2}\cdot a_{2}.\]根据题意，$a_{2k-1},a_{2k},a_{2k+1}$ 成等差数列，于是 $2a_{2k}=a_{2k-1}+a_{2k+1}$，即$$2=q_{k}+\dfrac{1}{q_{k-1}}.$$（i）设 $b_{k}=\dfrac{1}{q_{k}-1}$，则 $q_{k}=\dfrac{1}{b_{k}}+1$，于是\[2=\dfrac{1}{b_{k}}+1+\dfrac{b_{k-1}}{b_{k-1}+1},\]即$$b_{k}(b_{k-1}+1)=b_{k-1}+1+b_{k}b_{k-1},$$于是$$b_{k}-b_{k-1}=1,$$因此 $\left\{\dfrac{1}{q_{k}-1}\right\}$ 是公差为 $1$ 的等差数列．
（ii）原命题即证明 $\displaystyle 2n-2\leqslant \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}<2n-\dfrac{3}{2}$．
因为 $a_{1}=0$，$a_{2}=2$，所以 $d_{1}=2$，从而$$a_{3}=4 , q_{1}=\dfrac{a_{3}}{a_{2}}=2,$$因此$$q_{k}=\dfrac{k+1}{k},$$从而\[a_{2k}=k^{2}\cdot a_{2}=2k^{2}, a_{2k+1}=q_{k}\cdot a_{2k}=2k(k+1).\]情形一 当 $n$ 是奇数时，\[\begin{split}\sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}&=\dfrac{2^{2}}{2}+\dfrac{3^{2}}{2\cdot 1\cdot 2}+\cdots+\dfrac{(n-1)^{2}}{2\cdot \left(\dfrac{n-1}{2}\right)^{2}}+\dfrac{n^{2}}{2\cdot \dfrac{n-1}{2}\cdot \left(\dfrac{n-1}{2}+1\right)}\\&=n-1+2\left[\dfrac{3^{2}}{2\cdot 4}+\dfrac{5^{2}}{4\cdot 6}+\cdots+\dfrac{n^{2}}{(n-1)(n+1)}\right],\end{split}\]显然$$\dfrac{n^{2}}{(n-1)(n+1)}>1,$$于是$$\displaystyle \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}\geqslant n-1+2\cdot \dfrac{n-1}{2}=2n-2.$$用数学归纳法证明右边．归纳基础 当 $n=3$ 时，$$\displaystyle \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}=\dfrac{17}{4}\in\left[4,\dfrac{9}{2}\right),$$所以命题成立．递推证明 假设当 $n=m$ 时，命题成立，即$$\displaystyle \sum\limits_{k=2}^{m}\dfrac{k^{2}}{a_{k}}<2m-\dfrac{3}{2}.$$则当 $n=m+2$ 时，\[\sum\limits_{k=2}^{m+2}\dfrac{k^{2}}{a_{k}}=\sum\limits_{k=2}^{m}\dfrac{k^{2}}{a_{k}}+2+\dfrac{(m+2)^{2}}{(m+1)(m+3)}<2(m+2)-\dfrac{3}{2}+\dfrac{(m+2)^{2}}{(m+1)(m+3)}-2,\]因为\[\dfrac{(m+2)^{2}}{(m+1)(m+3)}-2=\dfrac{m^{2}+4m+4-2\left(m^{2}+4m+3\right)}{(m+1)(m+3)}=\dfrac{-(m^{2}+4m+2)}{(m+1)(m+3)}<0,\]所以$$\displaystyle \sum\limits_{k=2}^{m+2}\dfrac{k^{2}}{a_{k}}<2(m+2)-\dfrac{3}{2},$$命题成立．
综上，当 $n$ 为奇数时，$\displaystyle 2n-2\leqslant \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}<2n-\dfrac{3}{2}$ 恒成立．情形二 当 $n$ 是偶数时，$$\displaystyle \sum\limits_{k=2}^{n}\dfrac {k^2}{a_k}=\sum\limits_{k=2}^{n+1}\dfrac {k^2}{a_k}-\dfrac {2n}{n+1},$$由情形一知，$$\displaystyle 2(n+1)-2-\dfrac {2n}{n+1}\leqslant \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}<2(n+1)-\dfrac{3}{2}-\dfrac {2n}{n+1},$$即\[2n-2\leqslant \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}<2n-\dfrac{3}{2},\]所以当 $n$ 是偶数时，$\displaystyle 2n-2\leqslant \sum\limits_{k=2}^{n}\dfrac{k^{2}}{a_{k}}<2n-\dfrac{3}{2}$ 恒成立．
综合以上两种情况，原不等式得证．
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