sn是公比为q的等比数列{an}的前n项之和,且sn\=0,sn,s2n-sn,s3n-s2n的公比是?要详解

Question

sn是公比为q的等比数列{an}的前n项之和,且sn\=0,sn,s2n-sn,s3n-s2n的公比是?
要详解

Answer 1

sn是等比数列的前n项和,
若q不等于1,则
sn=a1(q^n-1)/(q-1)
s2n=a1(q^(2n)-1)/(q-1)
s3n=a1(q^(3n)-1)/(q-1)
所以
s2n-sn=a1{[q^(2n)-1]-(q^n-1)}/(q-1)=a1*(q^n-1)q^n/(q-1)
s3n-s2n=a1{[q^(3n)-1]-[q^(2n)-1]}/(q-1)=a1*(q^n-1)q^(2n)/(q-1)
故很明显得出sn,s2n-sn,s3n-s2n,的公比为q^n
若q=1时,更容易证明上面数列的公比也为1,即也符合q^n,但不能用上式来求得.