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

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Answer 3

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