(an,an+1)在函数f(x)=x^2+4x+2的图像上
a(n+1)+2=(an+2)^2
设cn=an+2 c1=a1+2=3
cn=c(n-1)^2=c1^2^(n-1)=3^2^(n-1)
an=cn-2=3^2^(n-1)-2
1/(an+1)=1/(3^2^(n-1)-1) 1/(an+3)=1/(3^2^(n-1)+1)
bn是1/an+1与1/an+3的等差中项,
bn=[3^2^(n-1)]/[(3^2^n)-1]
设3^2^(n-1)=m
则bn=m/(m-1)(m+1)
=-2[1/(1-m)-1/(1+m)]
天呢!你这题太繁杂了,你自己算吧.