Question

数列{an}的前n项和为Sn,如果a1=1/3 sn=n+2/3 *an a9=?

Answer 1

sn=[(n+2)/3] an .(1)

=>S(n-1)=[(n+1)/3]a(n-1).(2)

(1)-(2)

Sn-S(n-1)=[(n+2)/3] an -[(n+1)/3]a(n-1)

=>an=[(n+2)/3] an -[(n+1)/3]a(n-1)

=>[(n+1)/3]a(n-1)={[(n+2)/3] -1}an =[(n-1)/3]an

=>an=[(n+1)/(n-1)a(n-1)

=>a2=3a1=1=3/3

a3=2a2=6/3

a4=(5/3)a3=10/3

a5=(6/4)a4=6/4*10/3=15/3

a6=7/5*a5=7=21/3

a7=8/6*7=28/3

.

a8=36/3=12

a9=45/3=15