Let $f\left(x\right)$ be a function such that; $f(x-1)+f(x+1)=\sqrt{3}f\left(x\right),\forall x\in R$. If $f\left(5\right)=100$, then the value of $\sum _{r=0}^{99}f\left(512r\right)$.

The given function is $f(x-1)+f(x+1)=\sqrt{3}f\left(x\right)$ and $\forall x\in R.$ Also, $f\left(5\right)=100$.

Assume $x\to x+1$,

$f\left(x\right)+f(x+2)=\sqrt{3}f(x+1) ....\left(1\right)$

Assume $x\to x-1$,

$f(x-2)+f\left(x\right)=\sqrt{3}f(x-1) ....\left(2\right)$

Add equation $\left(1\right)$ and $\left(2\right)$,

$f\left(x\right)+f(x+2)+f(x-2)+f\left(x\right)$

$=\sqrt{3}f(x-1)+\sqrt{3}f(x+1)f(x+2)+f(x-2)+2f\left(x\right)$

$=\sqrt{3}\left\{f\right(x-1)+f(x+1\left)\right\}f(x+2)+f(x-2)+2f\left(x\right)$

$=\sqrt{3}\left\{\sqrt{3}f\right(x\left)\right\}f(x+2)+f(x-2)=f\left(x\right)$

Assume $\mathrm{x}\to x+2$,

$f(x+4)+f\left(x\right)=f(x+2) ....\left(3\right)$

Assume $\mathrm{x}\to x-2$,

$f(x-4)+f\left(x\right)=f(x-2) .....\left(4\right)$

Add equation $\left(3\right)$ and $\left(4\right)$,

$f(x+4)+f\left(x\right)+f(x-4)+f\left(x\right)$

$=f(x-2)+f(x+2)f(x+4)+f(x-4)+2f\left(x\right)$

$=f\left(x\right)f(x+4)+f(x-4)=-f\left(x\right)$

Continuing in this manner,

$f(x+12)=f\left(x\right)$ can be obtained which is a periodic function and the period is $12.$

Consider the given question,

$\sum _{r=0}^{99}f(5+12r)=f\left(5\right)+f(5+12)+f(5+24)+\cdots 100$

$\sum _{r=0}^{99}f(5+12r)=100\times f\left(5\right)$

$\sum _{r=0}^{99}f(5+12r)=100\times 100$

$\sum _{r=0}^{99}f(5+12r)=10000$

Hence, the value of $\sum _{r=0}^{99}f\left(512r\right)$ is $10000$.

Therefore, the correct answer is option $\left(3\right)$.