Question

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

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

Answer 1

Answer: (A)

Given that, $f(1)=1$,
$f(2n)=f(n),f(2n+1)=\{f(n){\}}^2-2$
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, $f(1)+f(2)+\dots +f(25)$
$=1+1-1+1-1+\dots +(-1)=1$