Question

If $f(1)=1, f(2 n)=f(n) $ and $f(2 n+1)=\{f(n)\}^{2}-2$ for $n=1,2,3, \ldots$, then the value of $f(1)+f(2)+\ldots+f(25)$ is

• A. 1
• B. -15
• C. -17
• D. -1
• E. 13

Answer 1

Answer: (A)

Given that, $f(1)=1$,
$f(2 n) =f(n), f(2 n+1)=\{f(n)\}^{2}-2 $
Now, $ f(2) =f(2 \times 1)=f(1)=1 $
$f(3) =f(2 \times 1+1)$
$=\{f(1)\}^{2}-2=1-2=-1$
$ f(4) =f(2 \times 2)=f(2)=1,$
$ f(5) =f(2 \times 2+1)$
$ =\{f(2)\}^{2}-2=1-2=-1 $
and so on
Now, $ f(1)+ f(2)+\ldots+f(25) $
$=1+1-1+1-1+\ldots+(-1)=1 $