若无穷数列an满足a1=2,an+1=根号3an+2,且有极限,则liman=

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 17:31:10
若无穷数列an满足a1=2,an+1=根号3an+2,且有极限,则liman=
xQMN@,JnsM+@V &4& 5&^L+8 buw~,ɯ|mvłGxyGi ]MG>䏃`r~v:42p\EU``\ M^l @GT0쩆fT2VBPPKLafsѴ;ԲڡU |1@fXR,6li]<U,˸Vޮٜ/T/HT봖_F{s1M?xѱ+#Ny27

若无穷数列an满足a1=2,an+1=根号3an+2,且有极限,则liman=
若无穷数列an满足a1=2,an+1=根号3an+2,且有极限,则liman=

若无穷数列an满足a1=2,an+1=根号3an+2,且有极限,则liman=
A(n+1) = √3*An + 2
A(n+1) + c = √3*An + 2 + c = √3 * [An + (2+c)/√3 ]
令 c = (2+c)/√3,则 c = 2/(√3 -1) = √3 + 1
所以有:
A(n+1) + √3 + 1 = √3 (An + √3 + 1)
可见,{An+√3+1}就是一个等比数列,因此有:
An+√3+1 = (A1 + √3 + 1) * (√3)^(n-1) = (√3+2) * (√3)^(n-1)
所以,An = (√3+2)*(√3)^(n-1) - √3 - 1
limAn 为无穷大.