Question

設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n,已知a₁=1,a₂=6,a₃=11,且(5n-8)S_n+1-(5n+2)S_n=An+B,n=1,2,3,…,其中A、B為常數(shù).

(1)求A與B的值;

(2)證明數(shù)列{a_n}為等差數(shù)列;

(3)證明不等式＞1對(duì)任何正整數(shù)m、n都成立.

Answer 1

思路解析:本題主要考查等差數(shù)列的定義、通項(xiàng)公式及利用分析法來證明問題.

(1)解:由已知,得S₁=a₁=1,S₂=a₁+a₂=7,S₃=a₁+a₂+a₃=18.由(5n-8)S_n+1-(5n+2)S_n=An+B,

知

即

解得A=-20,B=-8.

(2)證明:由(1)得(5n-8)S_n+1-(5n+2)S_n=-20n-8.                                    ①

所以(5n-3)S_n+2-(5n+7)S_n+1=-20n-28.                                           ②

②-①得(5n-3)S_n+2-(10n-1)S_n+1+(5n+2)S_n=-20.                                   ③

所以(5n+2)S_n+3-(10n+9)S_n+2+(5n+7)S_n+1=-20.                                   ④

④-③得(5n+2)S_n+3-(15n+6)S_n+2+(15n+6)S_n+1-(5n+2)S_n=0.

因?yàn)閍_n+1=S_n+1-S_n,

所以(5n+2)a_n+3-(10n+4)a_n+2+(5n+2)a_n+1=0.

因?yàn)?5n+2)≠0,

所以a_n+3-2a_n+2+a_n+1=0.

所以a_n+3-a_n+2=a_n+2-a_n+1,n≥1.

又a₃-a₂=a₂-a₁=5,

所以數(shù)列{a_n}為等差數(shù)列.

(3)證明:由(2)可知,a_n=1+5(n-1)=5n-4,

要證＞1,

只要證5a_mn＞1+a_ma_n+.

因?yàn)閍_mn=5mn-4,a_ma_n=(5m-4)(5n-4)=25mn-20(m+n)+16,

故只要證5(5mn-4)＞1+25mn-20(m+n)+16+,

即只要證20m+20n-37＞,

因?yàn)?SUB>≤a_m+a_n=5m+5n-8＜5m+5n-8+(15m+15n-29)=20m+20n-3.

所以原命題得證.