Question

一道关于数列的题（急急急！！！！）

已知数列an.a(n+1)*根号an=10,令bn=lg (an +1)/an,求证数列bn是等比数列

Answer 1

a（n+1）×根号an=10
lg(an+1)+1/2lgan=lg10
lgan+1/2lga(n-1)=lg10
联立
lga(n+1)=1/2[lgan+lga(n-1)]
a(n+1)*a(n+1)=an*a(n-1)
b1=lg10
b2=-1/2lg10
设bn=-(1/2)^(n-1)lg10
当k=1,b1=lg10 成立
假设bk=-(1/2)^k*lg10=lga(k+1)-lgak
b(k+1)=lga(k+2)-lga(k+1)
=lga(k+2)-lga(k+1)+lgak-lgak
=lga(k+2)-bk-lgak
=lg{a(k+2)/ak}-bk
= lg[a(k+1)/a(k+2)] -bk :a(n+1)*a(n+1)=an*a(n-1)
=-b(k+1)-bk
bk=-2b(k+1)
b(k+1)=-1/2*[-(1/2)^k*lg10}
=-(1/2)^(k+1)*lg10
命题成立
所以数列bn是等比
bn=-(1/2)^(n-1)lg10=lg（a（n+1）/an）=lga(n+1)-lgan