If bn+1=(1/1-bn) for n ≥ 1 and b1=b3, then displaystyle∑r=12001 br2001 is equal to

Question

If bn+1=(1/1-bn) for n ≥ 1 and b1=b3, then displaystyle∑r=12001 br2001 is equal to (A) 2001 (B) -2001 (C) 0 (D) None of these

Tardigrade Admin · Accepted Answer

b2=(1/1-b1) b3=(1/1-b2)=(1/1-(1)1-b1)=(1-b1/-b1)=(b1-1/b1) b1=b3 ⇒ b12-b1+1=0 ⇒ b1=-ω ⇒ b2=(1/1+ω)=-ω displaystyle∑r=12001 br2001= displaystyle∑r=12001-ω2001 = displaystyle∑r=12001-1=-2001

Q. If $b_{n + 1} = \frac{1}{1 - b _{n}}$ for $n \geq 1$ and $b_{1} = b_{3}$ , then $r = 1 \sum 2001 b_{r}^{2001}$ is equal to

$2001$

$- 2001$

$0$

None of these

Q. If bn+1​=1−bn​1​ for n≥1 and b1​=b3​, then r=1∑2001​br2001​ is equal to

2001

−2001

0

None of these

b2​=1−b1​1​ b3​=1−b2​1​=1−1−b1​1​1​=−b1​1−b1​​=b1​b1​−1​ b1​=b3​⇒b12​−b1​+1=0 ⇒b1​=−ω⇒b2​=1+ω1​=−ω r=1∑2001​br2001​=r=1∑2001​−ω2001 =r=1∑2001​−1=−2001

Q. If $b_{n + 1} = \frac{1}{1 - b _{n}}$ for $n \geq 1$ and $b_{1} = b_{3}$ , then $r = 1 \sum 2001 b_{r}^{2001}$ is equal to

$2001$

$- 2001$

$0$