Let $a=\min \left\{x^2+2 x+3, x \in R\right\}$  and  $b=\lim\limits_{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2}$, then The value of $\sum\limits_{r=0}^n a^r b^{n-r}$ is:

$\frac{4^{n+1}-1}{3.2^n}$

$x^2+2 x+3=(x+1)^2+2 \geq 2$. So a = 2

$b=\lim\limits_{\theta \rightarrow 0} \frac{2 \sin ^2 \frac{\theta}{2}}{\theta^2}=\frac{1}{2}$

Now $\sum\limits_{r=o}^n a^r b^{n-r}=b^n+a b^{n-1}+a^n b^n+..... a^n$

$=\frac{b^n\left(1-\frac{a}{b}\right)^{n+1}}{1-\frac{a}{b}}=\frac{\left(\frac{1}{2}\right)^n\left(1-4^{n+1}\right)}{1-4}=\frac{4^{n+1}-1}{3.2^n}$

Hence (3) is the correct answer.