Question

等比数列｛an｝中,若a1＋a2＋a3＝7,a1a2a3＝8,且公比大于1,求数列的前数列n和sn

Answer 1

a1a3=a2^2

a1a2a3＝8 a2=2

a1＋a2＋a3＝7

a1+a3=5

a1*a3=4

a1=1 a3=4 q=2

a1=4 a3=1 q=1/2(舍)

Sn=a1(1-q^n)/(1-q)

=2^n-1