高二数列+不等式

来源：百度知道 编辑：UC知道 时间：2024/09/25 07:05:03
数列A（N+1)=F(A(N))，F(X)=(-3/2）X^2+X,求证A(N)<1/N+1.
用数学归纳法证明。

令A(N)<1/N+1 ，则
A(N+1)
= (-3/2)A(N)^2 + A(N)
< (-3/2)(1/N+1)^2 + 1/N+1
= (-3/2)/(N^2+2N+1) + (N+1)/(N^2+2N+1)
= [N-(1/2)]/(N^2+2N+1)
= (N+2)[N-(1/2)]/(N+2)(N^2+2N+1)
= [N^2+(3/2)N-1]/(N+2)(N^2+2N+1)
= [N^2+(3/2)N-1]/(N+2)[N^2+(3/2)N-1+(N/2+2)]
< [N^2+(3/2)N-1]/(N+2)[N^2+(3/2)N-1] ..........分母变小分式变大
= 1/N+2
即A(N+1) < 1/[(N+1)+1]

数列A（N+1)=F(A(N))，F(X)=(-3/2）X^2+X,求证A(N)<1/N+1