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-(n+1)a2n]1-a2n 1-a2