Question

Consider the sequence $a_n$ given by $a_1=\frac{1}{2},a_{n+1}=a_n^2+a_n$.
Let $S_n=\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_n+1}$. Then find the value of $\left[S_{2012}\right]$, where [.] denotes greatest integer function.

• A. $1$
• B. $\left[\frac{e}{2}\right]$
• C. $\left[e\right]$
• D. $[\pi -1]$

Answer 1

Answer: (A), (B)

The correct options are
A $1$
B $\left[\frac{e}{2}\right]$

$a_{n+1}=a_n^2+a_n=a_n(a_n+1)$
$\Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_n(a_n+1)}=\frac{1}{a_n}-\frac{1}{a_n+1}$
$\frac{1}{a_{n+1}}=\frac{1}{a_n}-\frac{1}{a_n+1}\Rightarrow \frac{1}{a_n+1}=\frac{1}{a_n}-\frac{1}{a_{n+1}}$
Now, $S_{2012}=\frac{1}{a_1+1}+\frac{1}{a_2+1}+\frac{1}{a_3+1}+.....+\frac{1}{a_{2012}+1}$
$=(\frac{1}{a_1}-\frac{1}{a_2})+(\frac{1}{a_2}-\frac{1}{a_3})+.....+(\frac{1}{a_{2012}}-\frac{1}{a_{2013}})$
$S_{2012}=\frac{1}{a_1}-\frac{1}{a_{2013}}$ ..... (i)
Now, for the given sequence, $a_1=\frac{1}{2}$
$\Rightarrow a_2=\frac{1}{2}+\frac{1}{4}=\frac{3}{4},a_3=\frac{3}{4}+\frac{9}{16}>1$.
So, all the remaining terms in the sequence will also be greater than 1.
$a_{2013}>1\Rightarrow \frac{1}{a_{2013}}<1$

Hence $\left[S_{2012}\right]=[2-$a number between $0$ and $1]=1$