每日一题[860]层层展开

如图,曲线$y = \sqrt x $上的点${P_i}\left( {i = 1, 2, \cdots , n, \cdots } \right)$与$x$轴正半轴上的点${Q_i}$及原点$O$构成一系列正三角形${P_i}{Q_{i-1}}{Q_i}({Q_0} = O)$,记${a_n} = \left| {{Q_n}{Q_{n-1}}} \right|$.

(1)求${a_1}$的值;

(2)求数列$\{ {a_n}\} $的通项公式;

(3)求证:当$n \geqslant 2$时,$$\dfrac{1}{{{a_n}^2}} + \dfrac{1}{{{a_{n + 1}}^2}} + \cdots + \dfrac{1}{{{a_{2n}}^2}} < \dfrac{3}{2}.$$


cover

分析与解 (1)第一个正三角形的上顶点为直线$y = \sqrt 3 x$与抛物线$y = \sqrt x $在第一象限的交点,为$\left( {\dfrac{1}{3}, \dfrac{1}{{\sqrt 3 }}} \right)$,所以 ${a_1} = \dfrac{2}{3}$.

(2)设第$n$个正三角形的上顶点坐标为$\left( {y_n^2, {y_n}} \right)$,则左下顶点的坐标为$$\left( {y_n^2-\dfrac{{{y_n}}}{{\sqrt 3 }}, 0} \right),$$右下顶点的坐标为$$\left( {y_n^2 + \dfrac{{{y_n}}}{{\sqrt 3 }}, 0} \right).$$由于第$n$个正三角形的右下顶点与第$n + 1$个正三角形的左下顶点重合,于是$$y_n^2 + \dfrac{{{y_n}}}{{\sqrt 3 }} = y_{n+1}^2-\dfrac{{{y_{n + 1}}}}{{\sqrt 3 }},$$即$${y_{n + 1}}-{y_n} = \dfrac{1}{{\sqrt 3 }}.$$第一个正三角形的上顶点为$\left( {\dfrac{1}{3}, \dfrac{1}{{\sqrt 3 }}} \right)$,所以 $${y_1} = \dfrac{1}{{\sqrt 3 }},$$综合以上知${y_n} = \dfrac{n}{{\sqrt 3 }}$.

因此第$n$个正三角形的边长为$${a_n} = \dfrac{2}{{\sqrt 3 }}{y_n} = \dfrac{{2n}}{3}.$$(3)问题即证$$\dfrac{9}{4}\left[ {\dfrac{1}{{{n^2}}} + \dfrac{1}{{{{(n + 1)}^2}}} + \cdots + \dfrac{1}{{{{(2n)}^2}}}} \right] < \dfrac{3}{2},$$当$n = 2$时,不难验证$$\dfrac{9}{4}\left({\dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}}} \right)< \dfrac{3}{2},$$当$n \geqslant 3$时,\[\begin{split}&\dfrac{1}{{{n^2}}} + \dfrac{1}{{{{(n + 1)}^2}}} + \cdots + \dfrac{1}{{{{(2n)}^2}}}\\<&\dfrac{1}{{(n-1)n}} + \dfrac{1}{{n(n + 1)}} + \cdots + \dfrac{1}{{(2n-1)(2n)}}\\=& \dfrac{1}{{n-1}}-\dfrac{1}{{2n}} < \dfrac{1}{{n-1}} < \dfrac{2}{3},\end{split}\]所以$$\dfrac{9}{4}\left[{\dfrac{1}{n^2} + \dfrac{1}{{(n + 1)}^2}+ \cdots + \dfrac{1}{(2n)}^2} \right]< \dfrac{3}{2}$$成立.

此条目发表在每日一题分类目录,贴了标签。将固定链接加入收藏夹。

发表评论