8.在等比数列{a}中,a1+a2+…+a10=2,a11+a12+…+a30=12,求a31+a32+…+a60.

Question

8.在等比数列{a}中,a1+a2+…+a10=2,a11+a12+…+a30=12,求a31+a32+…+a60.

Answer 1

a1+a2+……a10=2,即 S10=a1(1-q^10)/(1-q)=2 (1),
a1+a2……+a10+a11+a12……a30=2+12=14 ,即S30= a1(1-q^30)/(1-q)=14 (2),
(2)/(1) 得 q^20+q^10+1=7 ,即 (q^10)^2+q^10-6=0 ,
所以 q^10=2 (已舍去负值)；
于是
a31+a32+……a60
=S60-S30
=a1(1-q^60)/(1-q)-a1(1-q^30)/(1-q)
=[a1(q-a^30)/(1-q)](1+q^30-1)
=S30(q^10)^3
=14*2^3=112.
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