数列{an}满足an=(n+2)(9/10)^n,当n为何值时,an有最大值?并求出此最大值

Question

数列{an}满足an=(n+2)(9/10)^n,当n为何值时,an有最大值?并求出此最大值

Answer 1

a1=27/10>0
an/a(n-1)
=(n+2)(9/10)^n/(n+1)(9/10)^(n-1)
=9(n+2)/10(n+1)
=(9n+18)/(10n+10)
=1+(8-n)/(10n+10),
当n<=7时,an/a(n-1)>1,a7>a6>a5>---->a1
n=8,an/a(n-1)=1,a7=a8
n>=9,an/a(n-1)<1,a8>a9>a10>---
所以当n=7,8时,an有最大值,此最大值为:
(9^8)/(10^7)