题目

已知｛an｝是整数组成的数列,a1=1,且点（根号an,an+1）（n∈N*）在函数y=x^2+1的图像上,an的通向公式n
若数列{bn}满足b1=1,bn+1=bn+2^an,求证：Bn*Bn+2

提问时间：2020-12-26

答案

(√an,a(n+1)) 在函数y=x^2+1
a(n+1) = an+1
a(n+1)-an = 1
{an}等差数列,d=1
an - a1 =n-1
an = n
b(n+1) =bn +2^an
= bn +2^n
bn - b(n-1) = 2^(n-1)
bn -b1 = 2+2^2+...+2^(n-1)
bn = 2^0+2^1+...+2^(n-1)
= 2^n - 1
(b(n+1))^2 = [2^(n+1) - 1]^2
= 2^(2n+2) - 2^(n+2) + 1
bn.b(n+2) = (2^n -1).(2^(n+2)-1)
= 2^(2n+2) - [2^n + 2^(n+2)] + 1
< 2^(2n+2) -2{√[2^n.2^(n+2)]} +1
=2^(2n+2) - 2^(n+2) +1
= (b(n+1))^2