Question

已知x1=2,xn+1=xn+1/xn,求X101的整数部分

Answer 1

x(n+1)=xn+1/xn
xn＞0
[x(n+1)]^2=(xn)^2+(1/xn)^2+2
(xn)^2=[x(n-1)]^2+[1/x(n-1)]^2+2
[x(n-1)]^2=[x(n-2)]^2+[1/x(n-2)]^2+2
……
(x4)^2=[x3]^2+[1/x3]^2+2
(x3)^2=[x2]^2+[1/x2]^2+2
(x2)^2=[x1]^2+[1/x1]^2+2
(xn)^2=[x(n-1)]^2+[1/x(n-1)]^2+2
=[x(n-2)]^2+[1/x(n-2)]^2+2+[1/x(n-1)]^2+2=[x(n-2)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*2
=[x(n-3)]^2+[1/x(n-3)]^2+2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*2=[x(n-3)]^2+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*3
……
=[x3]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-3)
=[x2]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-2)
=[x1]^2+[1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-1)
=2*(n+1)+[1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-1)
(xn)^2＞2*(n+1)
xn＞√[2*(n+1)]
又(1/xn)^2递减
[1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2
=1/4+4/25+100/841+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2
＜29/100+(100/841)(n-3)
=-5611/84100+(100/841)n
所以(xn)^2＜2*(n+1)-5611/84100+(100/841)n
√[2*(n+1)] ＜xn＜√[2*(n+1)-5611/84100+(100/841)n]
√204 ＜x101＜√215.94
14.28 ＜x101＜14.69
x101整数部分14