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

JEE Maths question with a full step-by-step solution.

Question

The sequence

\{x_n\}

is defined by

x_{k+1}=x_k^2+x_k

and

x_1=\frac12

. Then

\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

0

2

4

DNone of thesecorrect

Solution

Step 1:

x_{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

\frac{1}{x_k+1}=\frac{1}{x_k}-\frac{1}{x_{k+1}}.

Step 2: Sum from

k=1

100

; it telescopes, with

x_1=\frac12\Rightarrow\frac{1}{x_1}=2

\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

1

, so

0<\dfrac{1}{x_{101}}<1\Rightarrow

the sum lies in

(1,2)

. Its greatest integer is

1

— not among (1)–(3). Correct answer: (4)

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