Question

1.在各项都为正数的等比数列{an}中,a1=3.前三项的和为21,则a3+a4+a5=?

Answer 1

设比为q

a1=3,a2=3q a3=3q^2

所以 3+3q+3q^2=21

1+q+q^2=7

因为数列各项均为正数,所以q=2

a3+a4+a5=3q^2+3q^3+3q^4=3q^2*(1+q+q^2)=21q^2=84

2.a1*a2*a3……a30

=a1*a1*2*a1*2^2……*a1^29

=a1^30*2^(1+2+3+……+30)

=a1^30*2^435=2^30

所以a1^30=2^(-405) a1^10=2^(-135)

a3*a6*a9……*a30

=a1*2^2*a1*2^5*a1*2^8……*a1*2^29

=a1^10*2^(2+5+8+……+29)

=a1^10*2^155

=2^(-135)*2^155

=2^20