第1个回答 2009-10-21
an=a²(n-1)+a(n-1)
an=(a(n-1)+1/2)²-1/4
an+1/2=(a(n-1)+1/2)²+1/4
bn=an+1/2,b1=a1+1/2=1
bn=b²n+1/4
设x+1/kx=b1,(x+1/kx)²+1/4=x²+1/(kx)²+2/k+1/4,
令2/k=-1/4,k=-8,x-1/8x=1,
既有b2=b²1+1/4=x²+(1/kx)²,b3=b²2+1/4=(x²)²+((1/kx)²)²
x+1/(-8x)=1的,8x²-8x-1=0,x=(1+根号6)/2,x=(1-根号6)/2
b1=1=(1+根号6)/2+1/8*((1-根号6)/2)=(1+根号6)/2-(1+根号6)/20
b2=b²1+1/4=x²+(1/kx)²
b3=b²2+1/4=(x²)²+((1/kx)²)²
b4=b²3+1/4=(x^(2*2*2))+(1/kx)^(2*2*2)
...
bn=b(n-1)+1/4=(x^(2*2*2...*2)((n-1)个2)+(1/kx)^(2*2*...*2)((n-1)个2)
bn=x^(2^(n-1))+(1/kx)^(2^(n-1))
an=x^(2^(n-1))+(1/kx)^(2^(n-1))-1/2
将x=(1+根号6)/2(或x=(1-根号6)/2),任意一个),k=-8带入即可