每日一题[672]纠缠不清

等差数列$\{a_n\}$各项均为正整数,满足$a_1a_2-8a_1+a_2-13=0$,数列$\{b_n\}$满足$b_n=n^2$($n\in\mathcal N^*$),数列$\{a_n\}$与$\{b_n\}$所有公共项从小到大排列得到数列$\{c_n\}$,$S_n$是数列$\left\{\sqrt{1+\dfrac{1}{b_n}+\dfrac{1}{b_{n+1}}}\right\}$的前$n$项和,则$(4S_n)^2-c_{2n-1}$的最大值为_______.


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分析与解 根据题意,有$$(a_1+1)(a_2-8)=5,$$于是$a_1=4$,$a_2=9$,从而$$a_n=5n-1,n\in\mathcal N^*.$$考虑到当$n$模$5$的余数为$0,1,2,3,4$时,$n^2$模$5$的余数分别为$0,1,4,4,1$.于是$\{c_n\}$为模$5$余$2$或$3$的正整数从小到大排列得到的数列,进而$c_{2n-1}$表示第$n$个模$5$余$2$的数,即$$c_{2n-1}=(5n-3)^2,n\in\mathcal N^*.$$而$$\sqrt{1+\dfrac{1}{b_n}+\dfrac{1}{b_{n+1}}}=\sqrt{1+\dfrac 1{n^2}+\dfrac{1}{(n+1)^2}}=1+\dfrac 1n-\dfrac{1}{n+1},$$于是$$S_n=n+1-\dfrac{1}{n+1},n\in\mathcal N^*.$$这样就有\[\begin{split} (4S_n)^2-c_{2n-1}&=16(n+1)^2+\dfrac{16}{(n+1)^2}-(5n-3)^2-32\\&=-9n^2+62n+\dfrac{16}{(n+1)^2}-25\\&=-9(n+1)^2+80(n+1)+\dfrac{16}{(n+1)^2}-96.\end{split} \]设$\varphi(x)=-9x^2+80x+\dfrac{16}{x^2}$,则其导函数$$\varphi'(x)=-18x+80-\dfrac{32}{x^3},$$于是当$x\geqslant 2$时,$\varphi(x)$先增后减,且极大值点位于区间$(4,5)$.当$n=3$时,原式值为$81$;当$n=4$时,原式值为$\dfrac{1991}{25}$,因此所求的最大值为$81$.

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