Question

设数列{an}满足a1=1，且an+1=1+2/an（a>=1),求an

Answer 1

解：a(n+1)+1=1+2/an+1=2(an+1)/an
a(n+1)-2=1+2/an-2=-(an-2)/an
两式左右相除得[a(n+1)+1]/[a(n+1)-2]=-(an+1)/(an-2)
故{(an+1)/(an-2)}是首项为(a1+1)(a1-2)=-2 公比为-1的等比数列
则](an+1)/(an-2)=2*（-1）^2
解得an=[1+4(-1)^n]/[2(-1)^n-1]

Answer 2

a(n+1)=1+2/a(n)
a(n+1)+1=2+2/a(n),取倒数
1/[a(n+1)+1] =a(n)/[2a(n)+2]=[a(n)+1-1]/[2a(n)+2]=-(1/2){1/[a(n)+1]}+(1/2)
∴1/[a(n+1)+1]={[-(1/2)]^n}{1/[a(1)+1]}+(1/2)-(1/4)+(1/8)-...+(-1)^(n+1)(1/2^n)
n为偶数 ，1/[a(n)+1]=(1/2)^(n-1)-(1/2)^(n-2)+...+(1/2)-(1/2)^n=(1/4)+(1/16)+...+(1/4)^(n/2)=(2^n-1)/[3*(2^n)]
n为奇数 ，1/[a(n)+1]=-(1/2)^(n-1)+(1/2)^(n-2)-...+(1/2)+(1/2)^n=(1/4)+(1/16)+...+(1/4)^[(n-1)/2]+(1/2)^n=[2^(n-1)-1]/{3*[2^(n-1)]}+(1/2)^n=(2^n+1)/[3*(2^n)]
∴a(n)=[3*(2^n)]/(2^n-1)-1=[2^(n+1)+1]/(2^n-1) n为偶数
a(n)=[3*(2^n)]/(2^n+1)-1=[2^(n+1)-1]/(2^n+1) n为奇数

Answer 3