设{an}是正项等比数列,令Sn=lga1+lga2+...+laan.如果存在互异正整数m,n,使得Sm=Sn,则Sm+n=

设{an}是正项等比数列,令Sn=lga1+lga2+...+laan.如果存在互异正整数m,n,使得Sm=Sn,则Sm+n=
设{an}是正项等比数列,令Sn=lga1+lga2+...+laan.如果存在互异正整数m,n,使得Sm=Sn,则Sm+n=

设{an}是正项等比数列,令Sn=lga1+lga2+...+laan.如果存在互异正整数m,n,使得Sm=Sn,则Sm+n=

数列为正项等比数列,则首项a1>0,公比q>0
Sn=lga1+lga2+...+lgan
=lg(a1×a2×...×an)
=lg[a1×a1q×a1q²×...×a1q^(n-1)]
=lg[a1ⁿ×q^(1+2+...+n-1)]
=lg[a1ⁿ×q^[n(n-1)/2]]
=lg(a1ⁿ)+lg[q^[n(n-1)/2]]
=nlga1 +[n(n-1)/2]lgq
Sm=Sn
mlga1+[m(m-1)/2]lgq=nlga1+[n(n-1)/2]lgq
2(m-n)lga1+[m(m-1)-n(n-1)]lgq=0
2(m-n)lga1+(m²-m-n²+n)lgq=0
2(m-n)lga1+[(m+n)(m-n)-(m-n)]lgq=0
2(m-n)lga1+(m-n)(m+n-1)lgq=0
m、n为互异正整数,m-n≠0,等式两边同除以m-n
2lga1+(m+n-1)lgq=0
lga1=-[(m+n-1)/2]lgq
S(m+n)=(m+n)lga1 +[(m+n)(m+n-1)/2]lgq
=-(m+n)[(m+n-1)/2]lgq+[(m+n)(m+n-1)/2]lgq
=0