问题 解答题

已知:f(x)=logax(0<a<1).若数列{an} 使得2,f(a1),f(a2),…,f(an),2n+4(n∈N*)成等差数列.

(1)求数列{an}的通项;

(2)设bn=anf(an),若{bn}的前n项和为Sn,求Sn

答案

(1)2n+4=2+(n+1)d,∴d=2,

f(an)=2+2n=logaan,∴an=a2n+2

(2)bn=(2n+2)a2n+2,Sn=4a4+6a6+…+(2n+2)a2n+2,①

a2Sn=4a6+6a8+…+2na2n+2+(2n+2)a2n+4,②

②-①,整理,得Sn=

2a4
1-a2
[
1-a2n
1-a2
+1-(n+1)a2n]

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