Question

If $b_{n+1}=\frac{1}{1-b_n}$ for $n \geq 1$ and $b_1=b_3$, then $\displaystyle\sum_{r=1}^{2001} b_r^{2001}$ is equal to

• A. $2001$
• B. $-2001$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$b_2=\frac{1}{1-b_1}$
$b_3=\frac{1}{1-b_2}=\frac{1}{1-\frac{1}{1-b_1}}=\frac{1-b_1}{-b_1}=\frac{b_1-1}{b_1}$
$b_1=b_3 \Rightarrow b_1^2-b_1+1=0$
$\Rightarrow b_1=-\omega \Rightarrow b_2=\frac{1}{1+\omega}=-\omega$
$ \displaystyle\sum_{r=1}^{2001} b_r^{2001}=\displaystyle\sum_{r=1}^{2001}-\omega^{2001}$
$=\displaystyle\sum_{r=1}^{2001}-1=-2001$