每日一题[1487]等比放缩

若一个数列各项取倒数后按原来的顺序构成等差数列,则称这个数列为调和数列.已知数列 $\{a_{n}\}$ 是调和数列,对于各项都是正数的数列 $\{x_{n}\}$,满足 $x_{n}^{a_{n}}=x_{n+1}^{a_{n+1}}=x_{n+2}^{a_{n+2}}$($n\in\mathbb N^{*}$).

1、求证:数列 $\{x_{n}\}$ 是等比数列.

2、把数列 $\{x_{n}\}$ 中所有项按如图所示的规律排成一个三角形数表,当 $x_{3}=8$,$x_{7}=128$ 时,求第 $m$ 行各数的和; \[\begin{array}{cccc}\\ x_{1}&&&\\ x_{2}&x_{3}&&\\ x_{4}&x_{5}&x_{6}&\\ x_{7}&x_{8}&x_{9}&x_{10}\\ \cdots&\cdots&&\\ \end{array}\]

3、对于 $(2)$ 中的数列 $\{x_{n}\}$,证明:$\dfrac{n}{2}-\dfrac{1}{3}<\dfrac{x_{1}-1}{x_{2}-1}+\dfrac{x_{2}-1}{x_{3}-1}+\cdots+\dfrac{x_{n}-1}{x_{n+1}-1}<\dfrac{n}{2}$.

解析

1、因为$$x_{n}^{a_{n}}=x_{n+1}^{a_{n+1}}=x_{n+2}^{a_{n+2}},$$且数列 $\{x_{n}\}$ 中各项都是正数,所以\[a_{n}\lg x_{n}=a_{n+1}\lg x_{n+1}=a_{n+2}\lg x_{n+2}.\]设\[a_{n}\lg x_{n}=a_{n+1}\lg x_{n+1}=a_{n+2}\lg x_{n+2}=p, \]则\[\dfrac{p}{a_{n}}=\lg x_{n},\dfrac{p}{a_{n+3}}=\lg x_{n+1},\dfrac{p}{a_{n+2}}=\lg x_{n+2},\] 因为数列 $\{a_{n}\}$ 是调和数列,故 $a_{n}\ne 0$,\[\dfrac{2}{a_{n+1}}=\dfrac{1}{a_{n}}+\dfrac{1}{a_{n+2}},\]所以$$\dfrac{2p}{a_{n+1}}=\dfrac{p}{a_{n}}+\dfrac{p}{a_{n+2}},$$代入得$$2\lg x_{n+1}=\lg x_{n}+\lg x_{n+2},$$即\[\lg x_{n+1}^{2}=\lg (x_{n}x_{n+2}),\]故 $x_{n+1}^{2}=x_{n}x_{n+2}$,所以数列 $\{a_{n}\}$ 是等比数列.

2、设 $\{x_{n}\}$ 的公比为 $q$,则 $x_{3}q^{4}=x_{2}$,即$$8q^{4}=128.$$由于 $x_{n}>0$,故 $q=2$,于是\[x_{n}=x_{3}q^{n-3}=8\times 2^{n-3}=2^{n}.\]注意到第 $n$ 行共有 $n$ 个数,所以三角形数表中第 $1$ 行至第 $m-1$ 行共含有$$1+2+3+\cdots+(m-1)=\dfrac{m(m-1)}{2}$$个数,因此第 $m$ 行第 $1$ 个数是数列 $\{x_{n}\}$ 中第 $\dfrac{m(m-1)}{2}+1=\dfrac{m^{2}-m+2}{2}$ 项. 故第 $m$ 行各数的和为$$S_{m}=\dfrac{2^{\frac{m^{2}-m+2}{2}(2^{m}-1)}}{2-1}=2^{\frac{m^{2}+m+2}{2}}(2^{m}-1).$$

3、因为 $x_{n}=2^{n}$,所以\[\dfrac{x_{k}-1}{x_{k+1}-1}=\dfrac{2^{k}-1}{2^{k+1}-1}=\dfrac{2^{k}-1}{2\left(2^{k}-\dfrac{1}{2}\right)}<\dfrac{1}{2}.\]所以\[\dfrac{x_{1}-1}{x_{2}-1}+\dfrac{x_{2}-1}{x_{3}-1}+\cdots+\dfrac{x_{n}-1}{x_{n+1}-1}<\dfrac{1}{2}+\dfrac{1}{2}+\cdots+\dfrac{1}{2}=\dfrac{n}{2}.\]因为\[\begin{split}\dfrac{x_{k}-1}{x_{k+1}-1}&=\dfrac{2^{k}-1}{2^{k+1}-1}\\&=\dfrac{1}{2}-\dfrac{1}{2(2^{k+1}-1)}\\&=\dfrac{1}{2}-\dfrac{1}{3\cdot 2^{k}+2^{k}-2}\\ &\geqslant \dfrac{1}{2}-\dfrac{1}{3}\cdot \dfrac{1}{2^{k}}(k=1,2,3,\cdots,n),\end{split}\]设题中和式为 $M$,则\[\begin{split}M&=\dfrac{x_{1}-1}{x_{2}-1}+\dfrac{x_{2}-1}{x_{3}-1}+\cdots+\dfrac{x_{n}-1}{x_{n+1}-1}\\&\geqslant \left(\dfrac{1}{2}+\dfrac{1}{2}+\cdots+\dfrac{1}{2}\right)-\dfrac{1}{3}\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{2}+\cdots+\left(\dfrac{1}{2}\right)^{n}\right]\\&=\dfrac{n}{2}-\dfrac{1}{3}\cdot \dfrac{\dfrac{1}{2}\left[1-\left(\dfrac{1}{2}\right)^{n}\right]}{1-\dfrac{1}{2}}\\&=\dfrac{n}{2}-\dfrac{1}{3}\cdot \left[1-\left(\dfrac{1}{2}\right)^{n}\right]\\ &>\dfrac{n}{2}-\dfrac{1}{3}.\end{split}\]因此\[\dfrac{n}{2}-\dfrac{1}{3}<\dfrac{x_{1}-1}{x_{2}-1}+\dfrac{x_{2}-1}{x_{3}-1}+\cdots+\dfrac{x_{n}-1}{x_{n+1}-1}<\dfrac{n}{2}.\]

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