Question

已知等比数列中，前n项和Sn=54，S2n=60,则S3n等于？

详细过程。。。

Answer 1

对等比数列有个规律，就是Sn,S2n-Sn,S3n-S2n也成等比数列…S2n-Sn=6,S3n-S2n=2/3,S3n=182/3

也可以这样做
Sn=a1(1-q^n)/(1-q)=54
S2n=a1(1-q^2n)/(1-q)=60
S2n/Sn=(1-q^2n)/(1-q^n)=10/9
1+q^n=10/9
q^n=1/9
S3n=a1(1-q^3n)/(1-q)=(1+q^n+q^2n)Sn=(1+1/9+1/81)Sn
=91*54/81
=182/3

Answer 2

Sn (S2n-Sn) (S3n-S2n) 是等差数列
很简单就算出来了
记住这个规律
很有用的

Answer 3

Sn=a1(q^n-1)/(q-1)=48
S2n=a1(q^2n-1)/(q-1)=a1(q^n-1)(q^n+1)=60
相除
q^n+1=5/4
q^n=1/4
a1/(q-1)=48/(q^n-1)=-64

S3n=a1(q^3n-1)/(q-1)
=[a1/(q-1)]*(q^n-1)(q^2n+q^n+1)
=-64*(1/4-1)(1/16+1/4+1)
=63