Question

The sequence $a_n$ is defined by $a_1=\frac{1}{2}, a_{n+1}=a_n^2+a_n$. Also, $s=\frac{1}{a_1+1}+\frac{1}{a_2+1}+.....+\frac{1}{a_{100}+1}$ then $[S]$ (where [.] denotes the greatest integer function) is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

Answer 2

Given the recurrence $a_{n+1} = a_n(a_n+1)$, we can rewrite the fraction as:

$\frac{1}{a_{n+1}} = \frac{1}{a_n(a_n+1)}$

Using partial fractions:

$\frac{1}{a_n(a_n+1)} = \frac{1}{a_n} - \frac{1}{a_n+1}$

Thus,

$\frac{1}{a_n+1} = \frac{1}{a_n} - \frac{1}{a_{n+1}}$

The sum becomes telescopic:

$S = \sum_{n=1}^{100}\frac{1}{a_n+1} = \left(\frac{1}{a_1} - \frac{1}{a_2}\right)+\left(\frac{1}{a_2} - \frac{1}{a_3}\right)+ \cdots +\left(\frac{1}{a_{100}} - \frac{1}{a_{101}}\right)$

Telescoping leaves us with:

$S = \frac{1}{a_1} - \frac{1}{a_{101}}$

Given $a_1=\frac{1}{2}$, we have $\frac{1}{a_1}=2$. Since the sequence grows very fast, $\frac{1}{a_{101}}$ is extremely small. Therefore,

$S \approx 2 - \epsilon$ ($\epsilon$ is a very small positive number).

Thus, the greatest integer $[S]$ is $[S] = 1$.