Question

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

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

Answer 1

The correct option is A $1$

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

Now, $f\left(2\right)=f(2\times 1)=f\left(1\right)=1$,

$f\left(3\right)=f(2\times 1+1)$
$={\left\{f\left(1\right)\right\}}^2-2=1-2=-1$,
$f\left(4\right)=f(2\times 2)=f\left(2\right)=1$
$f\left(5\right)=f(2\times 2+1)$

$={\left\{f\left(2\right)\right\}}^2-2=1-2=-1$
and so on

Now, $f\left(1\right)+f\left(2\right)+\cdots +f\left(25\right)=1+1-1+1-1+\cdots +(-1)=1$