Question

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

Ans: (C)
Hint : $a=\min \left\{(x+1)^{2}+2\right\}=2$
$$\mathrm{b}=\operatorname{lt}_{\theta \rightarrow 0} \frac{2 \sin ^{2} \frac{\theta}{2}}{4\left(\frac{\theta}{2}\right)^{2}}=\frac{1}{2}, \mathrm{a}^{\mathrm{r}} \mathrm{b}^{n-r}=\frac{2^{r}}{2^{n-t}}=2^{2 r-n}=\frac{4^{r}}{2^{n}}, \sum_{t=0}^{n} \mathrm{a}^{r} \cdot \mathrm{b}^{n-r}=\frac{1}{2^{n}} \sum_{r=0}^{n} 4^{r}=\frac{1}{2^{n}}\left(\frac{1-4^{n+1}}{1-4}\right)=\frac{4^{n+1}-1}{3 \times 2^{n}}$$

Answer 2

online cialis <a href="https://ordergnonline.com/">brand name cialis</a> natural ed pills

Answer 3

order tadalafil 5mg generic <a href="https://ordergnonline.com/">tadalafil 5mg price</a> cheap erectile dysfunction pills online