Question

Given f(1) = 2 and f(n + 1) $=\frac{f\left(n\right)-1}{f\left(n\right)+1}\forall nϵN$ then which of the following is/ are correct?

• A. $f\left(2015\right)=-\frac{1}{2}$
• B. $\left(f\right(2012){)}^{f\left(2013\right)}$
• C. f(1001) = 2
• D. f(2015) = – 3

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$f\left(2015\right)=-\frac{1}{2}$

B

$\left(f\right(2012){)}^{f\left(2013\right)}$

C

f(1001) = 2

$\because f(n+1)=\frac{f\left(n\right)-1}{f\left(n\right)+1}=\frac{\frac{f(n-1)-1}{f(n-1)+1}-1}{\frac{f(n-1)-1}{f(n-1)+1}+1}=\frac{-1}{f(n-1)}\therefore f(n+1)=-\frac{-1}{f(n-1)}\cdots \left(i\right)$
In a similar way $f(n-1)=-\frac{-1}{f(n-3)}$ . . . (ii)
from (i) and (ii)
$f(n+1)=-\frac{1}{\frac{-1}{f(n-3)}}=f(n-3)orf(n+4)=f\left(n\right)\forall nϵN$
hence f(n) is a periodic sequence with period 4
Put n = 1, 2, 3
we get f(2) $=\frac{1}{3},f\left(3\right)=-\frac{1}{2},f\left(4\right)=-3$
Now f(2012) = f(4) = –3
f(2013) = f(1) = 2
f(2015) = f(3) = $-\frac{1}{2}$
f(1001) = f(1) = 2