Question

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

• A. $\frac{2^{n+1}-1}{3 \times 2^{n}}$
• B. $\frac{2^{n+1}+1}{3 \times 2^{n}}$
• C. $\frac{4^{n+1}-1}{3 \times 2^{n}}$
• D. None of these

Answer 1

Answer: (C)

$a=\min \left\{x^{2}+2 x+3, x \in R\right\}$
$=\min \left\{(x+1)^{2}+2, x \in R\right\}=2$ and
$b=\displaystyle \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}}$
$=\displaystyle \lim _{\theta \rightarrow 0} \frac{(1-\cos \theta)(1+\cos \theta)}{\theta^{2}(1+\cos \theta)}=\frac{1}{2}$
$\therefore \displaystyle \sum_{r=0}^{n} a^{r} \cdot b^{n-r}=b^{n} \displaystyle \sum_{r=0}^{n}\left(\frac{a}{b}\right)^{r}=\left(\frac{1}{2}\right)^{n} \sum_{r=0}^{n}(4)^{r}$
$=\frac{1}{2^{n}}\left(1+4+4^{2}+\ldots+4^{n}\right)$
$=\frac{1}{2 n} \cdot 1 \cdot\left(\frac{4^{n+1}-1}{4-1}\right)$
$=\frac{4^{n+1}-1}{3 \cdot 2^{n}}$