1、
把(√an,a(n+1))代入y=x²+1,得
a(n+1)=an + 1
即数列{an}是以1为首项,1为公差的等差数列.
∴an = n
2、
∵b(n+1)=bn+2^an
∴b(n+1) - bn = 2^an = 2^n
∴有：
bn - b(n-1) = 2^(n-1)
b(n-1) - b(n-2) = 2^(n-2)
b(n-2) - b(n-3) = 2^(n-3)
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b3 - b2 = 2²
b2 - b1 = 2
全加,得
bn - a1 = 2 + 2² + 2³ + …… + 2^(n-1) = 2^n - 2
∴bn = 2^n - 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^(n+2) + 1 - 2^n
=[2^(n+1)]² - 2×2^(n+1) + 1 -2^n
=[2^(n+1) - 1]² - 2^n
≤[2^(n+1) - 1]²
＝b²(n+1)