设等比数列{an}的前n项的和为Sn,前n项的倒数之和为Tn,则Sn/Tn=
问题描述：
设等比数列{an}的前n项的和为Sn,前n项的倒数之和为Tn,则Sn/Tn=
答
a1的平方*q(n-1)次方
答
设等比数列{an}的公比为q
侧：S n=a1(q的n次方-1)/(q-1)
Tn=1/a1+1/a2+,=1/a1[((1/q)的n次方-1)/(1/q-1)
= [(q的n次方-1)/(q-1)]/[ a1• q的(n-1`)次方]
=[(q的n次方-1)/(q-1)]/(an)
故：Sn/Tn=a1•an