数列{an}中有a1=1,a(n+3)= an+3,a(n+2)≥an+2 (n∈N*) (1) 求a7,a5,a3,a6 ;

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 07:30:00
数列{an}中有a1=1,a(n+3)= an+3,a(n+2)≥an+2 (n∈N*) (1) 求a7,a5,a3,a6 ;
xOJ0Ưec*%6`#xxRt) Wlt7mi[N2u'{_|jm^us]^\g aX!mWP(k6]w~QZ̵qtV]\XuuSZi'4pO1ac1/ߍNޟk,J˨t@DLHM|C !9Ed KnL?P2I~^mw ])^Wnj 

数列{an}中有a1=1,a(n+3)= an+3,a(n+2)≥an+2 (n∈N*) (1) 求a7,a5,a3,a6 ;
数列{an}中有a1=1,a(n+3)= an+3,a(n+2)≥an+2 (n∈N*) (1) 求a7,a5,a3,a6 ;

数列{an}中有a1=1,a(n+3)= an+3,a(n+2)≥an+2 (n∈N*) (1) 求a7,a5,a3,a6 ;
a(3)>=a(1)+2
a(5)>=a(3)+2
a(7)>=a(5)+2
a(7)>=a(1)+6
a(7)=a(4)+3=a(1)+3+3=a(1)+6
所以全部取等
a(3)=a(1)+2
a(5)=a(3)+2
a(7)=a(5)+2
a(3)=3
a(5)=5
a(7)=7
a(3)>=a(1)+2
a(5)>=a(3)+2
a(7)>=a(5)+2
a(9)>=a(7)+2
a(11)>=a(9)+2
a(13)>=a(11)+2
a(15)>=a(13)+2
a(17)>=a(15)+2
a(19)>=a(17)+2
a(19)>=a(1)+18
a(19)=a(16)+3=a(13)+6=a(10)+9=a(7)+12=19
所以以上全取等
a(15)=a(12)+3=a(9)+6=a(7)+2=9
a(6)=a(9)-3=6