设{an}是正项等比数列,令Sn=lga1+lga2+…+lgan,n∈N*,若存在互异的正整数m,n,使得Sm=Sn,则Sm+n=_.
问题描述:
设{an}是正项等比数列,令Sn=lga1+lga2+…+lgan,n∈N*,若存在互异的正整数m,n,使得Sm=Sn,则Sm+n=______.
答
∵{an}是正项等比数列,设公比为q,
∴lgan+1-lgan=lgq
∴数列{lgan}为等差数列,
设公差为d
则Sm=mlga1+
,Sn=nlga1+m(m−1)d 2
n(n−1)d 2
∵Sm=Sn,
∴Sm-Sn=mlga1+
-nlga1-m(m−1)d 2
=(m-n)(lga1+n(n−1)d 2
d)=0m+n−1 2
∵m≠n
∴lga1+
d)=0m+n−1 2
∴Sm+n=(m+n)lga1+
=(m+n)(lga1+(m+n)(m+n−1)d 2
d)=0m+n−1 2
故答案为0.