数列｛log2（An-1)(n为正整数）为等差数列,a1=3,a3=9,求｛An}的通项公式10

题目

数列｛log2（An-1)(n为正整数）为等差数列,a1=3,a3=9,求｛An}的通项公式10
证明1/（a2-a1)+1/(a3-a2)+……1/[A(n+1)-An}

答案

令{bn}={log2 (an-1)}
已知数列{bn}={log2 (an-1)}为等差数列,且a1=3 a3=9
所以
b1=log2 (3-1)=log2(2)=1,
b2=log2 (9-1)=log2(8)=3,
公差d=3-1=2,
所以bn=1+(n-1)×2,bn=2n-1
所以
log2 (an-1)=2n-1
所以
(1).an =2^(2n-1)
(2).a1=2^(1)=2,a2=2^(3)=8,a3=2^5=32,.
an =2^(2n-1),a(n+1) =2^(2n+1)
∴1/（a2-a1)+1/(a3-a2)+…+1/a(n+1)-an
=1/(8-2)+1/(32-8)+.+1/[2^(2n+1)-2^(2n-1)]
=1/6+1/24+.+1/3×(2^(2n-1)
=1/3×2+1/3×2^3+.1/3×(2^(2n-1)