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=______.

∵{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+

m(m−1)d
2
,Sn=nlga1+
n(n−1)d
2

∵Sm=Sn
∴Sm-Sn=mlga1+
m(m−1)d
2
-nlga1-
n(n−1)d
2
=(m-n)(lga1+
m+n−1
2
d
)=0
∵m≠n
∴lga1+
m+n−1
2
d
)=0
∴Sm+n=(m+n)lga1+
(m+n)(m+n−1)d
2
=(m+n)(lga1+
m+n−1
2
d
)=0
故答案为0.