The sequence xk is defined by xk+1=xk2+xk and x1=(1/2) . Then [(1/x1+1)+(1/x2+1)+ ldots+(1/x100+1)] (where [.] denotes the greatest integer function) is equal to

Question

The sequence xk is defined by xk+1=xk2+xk and x1=(1/2) . Then [(1/x1+1)+(1/x2+1)+ ldots+(1/x100+1)] (where [.] denotes the greatest integer function) is equal to (A) 0 (B) 2 (C) 4 (D) 1

Tardigrade Admin · Accepted Answer

(1/xk+1)=(1/xk(xk+1))=(1/xk)-(1/xk+1) ⇒ (1/xk+1)=(1/xk)-(1/xk-1) ∴ (1/x1+1)+(1/x2+1)+ ldots+(1/x100+1)=(1/x1)-(1/x101) ∴ As 0< (1/x101)< 1 ∴ [(1/x1+1)+(1/x2+1)+ ldots+(1/x100+1)]=1

Q. The sequence {xk​} is defined by xk+1​=xk2​+xk​ and x1​=21​. Then [x1​+11​+x2​+11​+…+x100​+11​] (where [.] denotes the greatest integer function) is equal to

0

2

4

1

xk+1​1​=xk​(xk​+1)1​=xk​1​−xk​+11​ ⇒xk+1​1​=xk​1​−xk−1​1​ ∴x1​+11​+x2​+11​+…+x100​+11​=x1​1​−x101​1​ ∴ As 0<x101​1​<1 ∴[x1​+11​+x2​+11​+…+x100​+11​]=1

Q. The sequence ${x_{k}}$ is defined by $x_{k + 1} = x_{k}^{2} + x_{k}$ and $x_{1} = \frac{1}{2} .$ Then $[\frac{1}{x _{1} + 1} + \frac{1}{x _{2} + 1} + \dots + \frac{1}{x _{100} + 1}]$ (where [.] denotes the greatest integer function) is equal to