已知f(x)=logaX a大于0 且a不等于1设f(a1),f(a2),f(an)是首项4公差2的等差数列

1） f(an)=4+2(n-1)= 2n+2
log(a,an)=2n+2
an=a^(2n+2),a是常数
a(n+1) /an =a²
所以 ,{an}成等比数列；
2）a1 =根号2,a^4 =根号2,a=2^(1/8)
bn = an*f(an) = (2n+2)* a^(2n+2) = (2n+2)*2^[(n+1)/4]
利用错位相减法,
Sn=4*2^(2/4) +6*2^(3/4) +...+ (2n+2)*2^[(n+1)/4]
Sn *2^(1/4) = 4*2^(3/4) +...+ 2n *2^[(n+1)/4] + (2n+2)*2^[(n+2)/4]
两式相减,得
[1- 2^(1/4)]Sn =4*2^(2/4) +2*2^(3/4) +...+ 2*2^[(n+1)/4] - (2n+2)*2^[(n+2)/4]
所以,Sn=4*2^(2/4)+ 2*2^3/4 *[1-2^(n-1)/4]/[1-2^(3/4)]² - (2n+2)*2^[(n+2)/4]/[1-2^(3/4)]