Question

Let a = min$\left\{x^{2}+2x+3 : x \in R\right\}$ and $b=\lim\limits_{\theta \to 0} \frac{1-cos\,\theta}{\theta^{2}}$. Then $\sum\limits^{n}_{r - 0} a^{r}b^{n-r}$ is

• A. $\frac{2^{n+1}-1}{3.2^{n}}$
• B. $\frac{2^{n+1}+1}{3.2^{n}}$
• C. $\frac{4^{n+1}-1}{3.2^{n}}$
• D. $\frac{1}{2}\left(2^{n}-1\right)$

Answer 1

Answer: (C)

$a=min {(x+1)^2+2} = 2$
$\lim\limits_{\theta \to 0}\frac{2\,sin^{2} \frac{\theta}{2}}{4\left(\frac{\theta}{2}\right)^{2}} = \frac{1}{2}, a^{r}\,b^{n-r}= \frac{2^{r}}{2^{n-r}} = 2^{2r-n} = \frac{4^{r}}{2^{n}}, \sum\limits^{n}_{r - 0} a^{r}.b^{n-r} = \frac{1}{2^{n}} \sum\limits^{n}_{r - 0} 4^{r} = \frac{1}{2^{n}} \left(\frac{1-4^{n+1}}{1-4}\right) = \frac{4^{n+1}-1}{3\times2^{n}}$