每日一题[2119]两边夹

已知正整数数列 $\{a_n\}$ 满足:$a_4=16$,且\[\dfrac1{a_{n+1}^2}<\dfrac1{a_n}+\dfrac1{a_{n+1}}-\dfrac2{n(n+1)}<\dfrac1{a_n^2},\]求数列 $\{a_n\}$ 的通项公式.

答案    $a_n=n^2$($n\in\mathbb N^{\ast}$)

解析    根据题意,显然有 $\{a_n\}$ 单调递增,且有\[ \dfrac{1}{a_1^2}>\dfrac{1}{a_1}+\dfrac{1}{a_2}-1>\dfrac{1}{a_2^2}>\dfrac1{a_2}+\dfrac{1}{a_3}-\dfrac 13>\dfrac1{a_3^2}>\dfrac{1}{a_3}+\dfrac{1}{a_4}-\dfrac 16>\dfrac{1}{a_4^2},\]由 $a_4=16$,可得\[\dfrac1{a_3^2}>\dfrac{1}{a_3}-\dfrac 5{48}>\dfrac{1}{256}\implies a_3=9,\]进而\[\dfrac{1}{a_2^2}>\dfrac{1}{a_2}-\dfrac 29>\dfrac1{81}\implies a_2=4,\]进而\[\dfrac{1}{a_1^2}>\dfrac{1}{a_1}-\dfrac 34>\dfrac1{16}\implies a_1=1.\]接下来利用数学归纳法证明,若 $a_k=k^2$($k\geqslant 4$),那么 $a_{k+1}=(k+1)^2$.此时有\[\dfrac{1}{a_{k+1}^2}<\dfrac{1}{k^2}+\dfrac{1}{a_{k+1}}-\dfrac{2}{k(k+1)}<\dfrac1{k^4},\]解得\[\dfrac{k^4(k+1)}{k^3-k^2+k+1}<a_{k+1}<\dfrac{k^3+k^2+k\sqrt{k^4+2k^3-3k^2+4}}{2(k-1)},\]从而\[-\dfrac{1+3k+2k^2}{1+k-k^2+k^3}<a_{k+1}-(k+1)^2<\dfrac{2(k+1)(2k+1)}{k\sqrt{k^4+2k^3-3k^2+4}+k^3+k^2-2k-2},\]由于\[(k^3-k^2+k+1)-(2k^2+3k+1)=k^3-2k^2-2k>0,\]于是\[-1<-\dfrac{1+3k+2k^2}{1+k-k^2+k^3}<0,\]且\[0<\dfrac{2(k+1)(2k+1)}{k\sqrt{k^4+2k^3-3k^2+4}+k^3+k^2-2k-2}<\dfrac{2(k+1)(2k+1)}{2k^3}<1,\]因此 $a_{k+1}-(k+1)^2=0$,从而 $a_{k+1}=(k+1)^2$,命题得证. 综上所述,数列 $\{a_n\}$ 的通项公式为 $a_n=n^2$($n\in\mathbb N^{\ast}$).

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