Question

An odd function is symmetric about the vertical line $x=a(a>0)$ and if $\sum _{r=0}^{\infty }\left[f\right(1+4r){]}^r=8,$ then find the numerical value of $8f\left(1\right)$

Answer 1

$f(2a-x)=f\left(x\right)$
$\Rightarrow f(2a-x)=-f\left(x\right)$
$\because f$ is odd $\Rightarrow f(x+4a)=f\left(x\right)$
$\Rightarrow f$ is periodic with period $4a$.
$\Rightarrow f(1+4r)=f\left(1\right)$
Now, $\sum _{r=0}^{\infty }\left[f\right(1){]}^r=8\Rightarrow \frac{1}{1-f\left(1\right)}=8$
$\Rightarrow f\left(1\right)=\frac{7}{8}\Rightarrow 8f\left(1\right)=7$