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Greatest Integer of a Telescoping Sum 1/(xk+1) | JEE

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Question
The sequence {xn}\{x_n\} is defined by xk+1=xk2+xkx_{k+1}=x_k^2+x_k and x1=12x_1=\frac12. Then [1x1+1+1x2+1+⋯+1x100+1]\left[\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_{100}+1}\right], where [ ⋅ ][\,\cdot\,] is the greatest integer function, equals
A00
B22
C44
DNone of thesecorrect
Solution
Step 1: xk+1=xk(xk+1)⇒1xk+1=1xk−1xk+1x_{k+1}=x_k(x_k+1)\Rightarrow\dfrac{1}{x_{k+1}}=\dfrac{1}{x_k}-\dfrac{1}{x_k+1}, which rearranges to
1xk+1=1xk−1xk+1.\frac{1}{x_k+1}=\frac{1}{x_k}-\frac{1}{x_{k+1}}.
 Step 2: Sum from k=1k=1 to 100100; it telescopes, with x1=12⇒1x1=2x_1=\frac12\Rightarrow\frac{1}{x_1}=2:
∑k=11001xk+1=1x1−1x101=2−1x101.\sum_{k=1}^{100}\frac{1}{x_k+1}=\frac{1}{x_1}-\frac{1}{x_{101}}=2-\frac{1}{x_{101}}.
 Step 3: The terms grow fast past 11, so 0<1x101<1⇒0<\dfrac{1}{x_{101}}<1\Rightarrow the sum lies in (1,2)(1,2). Its greatest integer is 11 — not among (1)–(3). Correct answer: (4)
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