设公差为d
d=log2(a2 -1)-log2(a1-1)=log2(4)-log2(2)=2-1=1
log2(an -1)=log2(a1 -1) +(n-1)d=1+n-1=n
an -1=2^n
an=2^n +1
a(n+1)-an=2^(n+1)+1-2^n-1=2^n
1/(a2-a1)+1/(a3-a2)+...+1/[a(n+1)-an]
=1/2^1 +1/2^2 +...+1/2^n
=(1/2)[1-(1/2)^n]/(1-1/2)
=1- 1/2^n
你是不是写错了,a1 a2都有了,直接带入不就得了,结果是1/2,前面条件根本没用上
∵{log2 ((an)-1)}为等差数列,a1=3 a2=5
∴log2 ((an)-1)=n
an=2^n+1
a(n+1)-an=2^(n+1)-2^n=2^n
1/(a2-a1)+1/(a3-a2)+…+(a(n+1)-an)=1/2+1/4+1/8+...+1/2^n=1-0.5^n