已知函数f（x）=log以k为底x的对数（k为常数,k＞0,且k≠1）且数列{f（an）}是首相为4,

(1) f(an)=logk(an)
f(an-1)=logk(an-1)
f(an)-f(an-1)=2
logk(an)-logk(an-1)=2
logk(an/an-1)=2
an/an-1=k^2 k^2为常数
所以 数列{an}是等比数列
logka1=4 a1=k^4
q=k^2
an=k^4*(k^2)^(n-1)=k^(2n+2)
(2) k=根号2
a1=4 q=2 an=2^(n+1)
f(an)=2n+2
bn=2(n+1)*2^(n+1)=4*[(n+1)*2^(n)]
Sn=4[2*2^1+3*2^2+4^2^3+……+(n+1)*2^n]
2Sn=4[ 2*2^2+3*2^3+4*2^4+……+n*2^n+(n+1)*2^(n+1)] 相减
-Sn=4[4+2^2+2^3+2^4+……+2^n-(n+1)*2^(n+1)]
=4[2+2(1-2^n)/(1-2)-(n+1)*2^(n+1)]
Sn=n*2^(n+3)