Question

已知数列{an}满足a1=1/2,3(an+1-an)(an+1-an)=(1+an+1)(1-an+1),且（an+1）*an小于0，n属于正整数，求{an}

Answer 1

原式化为4*an+1^2+6*an+1*an+3*an^2-1=0
(2*an+1+1)(2an+1-1)+3*an*(2*an+1+1)=0
(2*an+1)(2*an+1+3*an-1)=0可得:
an+1=-1/2 舍去
2*an+1+3*an-1=0
an+1=-3/2*an+1/2两边同减去1/5得
an+1-1/5=-3/2*(an-1/5)
设bn=an-1/5
则b1=a1-1/5=3/10
bn=3/10*(-3/2)^(n-1)
an=3/10*(-3/2)^(n-1)+1/5

Answer 2

3(an+1-an)(an+1-an)=(1+an+1)(1-an+1)
即4(an+1)^2=3(an)^2+1
即4[(an+1)^2-1]=3[(an)^2-1]
即[(an+1)^2-1]/[(an)^2-1]=3/4
即{(an)^2-1}等比
(an)^2-1=[(a1)^2-1]*(3/4)^(n-1)=-(3/4)^n
所以(an)^2=1-(3/4)^n
又（an+1）*an<0 a1=1/2>0
所以an=[(-1)^(n-1)][1-(3/4)^n]^(1/2) (n属于N+)