高中数学吧 关注：321,395贴子：2,381,036
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bn=1/(an-1),b(n-1)=1/[a(n-1)-1],
bn-b(n-1)=[1/(an-1)]-1/[a(n-1)-1]=[a(n-1)-an]/[(an-1)][a(n-1)-1]
=[a(n-1)-an]/[(ana(n-1)-an-a(n-1)+1] 因(ana(n-1)=2a(n-1)-1
=[a(n-1)-an]/[2a(n-1)-1-an-a(n-1)+1]
=[a(n-1)-an]/[a(n-1)-an]=1
即,bn-b(n-1)=1
{bn}是等差
b1=1/(a1-1)=1/(a1-1)=-5/2
bn=b1+n-1=-5/2+n-1=(2n-7)/2
bn=1/(an-1)=(2n-7)/2
an=(2n-5)/(2n-7)
bn-b(n-1)=[1/(an-1)]-1/[a(n-1)-1]=[a(n-1)-an]/[(an-1)][a(n-1)-1]
=[a(n-1)-an]/[(ana(n-1)-an-a(n-1)+1] 因(ana(n-1)=2a(n-1)-1
=[a(n-1)-an]/[2a(n-1)-1-an-a(n-1)+1]
=[a(n-1)-an]/[a(n-1)-an]=1
即,bn-b(n-1)=1
{bn}是等差
b1=1/(a1-1)=1/(a1-1)=-5/2
bn=b1+n-1=-5/2+n-1=(2n-7)/2
bn=1/(an-1)=(2n-7)/2
an=(2n-5)/(2n-7)
