Question

求详细过程啊！！！速度采纳！！写出来更好！！

Answer 1

追问

公比是四分之一，a1是四分之一，为什么通向公式也是四分之一不是应该十六分之一的n-1次方吗……抱歉基础差…

追答

追问

感谢

Answer 2

(1)an是等比数列，a1=1/4 a=1/4
an=a1*q^(n-1)
an=(1/4)^n
(2)bn+2=3log(1/4)(1/4)^n=3n
bn=3n-2
(3) cn=(3n-2)*(1/4)^n
Sn=1*1/4+4*(1/4)^2+7*(1/4)^3+……+(3n-2)*(1/4)^n
1/4Sn= 1*(1/4)^2+4*(1/4)^3+7*(1/4)^4+……+(3n+1)*(1/4)^n+……+(3n-2)*(1/4)^(n+1)
相减
3/4Sn=1/4+3*(1/4)^2+3*(1/4)^3+……+3*(1/4)^n-(3n-2)*(1/4)^(n+1)
=1/4+3[(1/4)^2+(1/4)^3+……+(1/4)^n]-(3n-2)*(1/4)^(n+1)
=1/4+3[1/3(1-(1/4)^n)]-(3n-2)*(1/4)^(n+1)
=1/4+1-3*(1/4)^n-(3n-2)*(1/4)^(n+1)
=5/4-3*(1/4)^n-(3n-2)*(1/4)^(n+1)
Sn=5/3-(1/4)^(n-1)-4(3n-2)/3*(1/4)^(n+1)