Let a = min {x^{2} + 2x + 3 : x in R} and b = | vedclass

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(C) First, find $a = \min \{x^{2} + 2x + 3\}$. The minimum value of a quadratic $Ax^{2} + Bx + C$ is given by $\frac{4AC - B^{2}}{4A}$. Here $A=1, B=2

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$\frac{4^{n+1}-1}{3 \cdot 2^{n}}$

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(C) First, find $a = \min \{x^{2} + 2x + 3\}$. The minimum value of a quadratic $Ax^{2} + Bx + C$ is given by $\frac{4AC - B^{2}}{4A}$. Here $A=1, B=2, C=3$, so $a = \frac{4(1)(3) - (2)^{2}}{4(1)} = \frac{12 - 4}{4} = 2$. Next, find $b = \lim_{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^{2}}$. Using the identity $1 - \cos 	heta = 2 \sin^{2}(	heta/2)$, we get $b = \lim_{	heta ightarrow 0} \frac{2 \sin^{2}(	heta/2)}{	heta^{2}} = \lim_{	heta ightarrow 0} \frac{2 \sin^{2}(	heta/2)}{4(	heta/2)^{2}} = \frac{2}{4}(1)^{2} = \frac{1}{2}$. Now, evaluate the sum $S = \sum_{r=0}^{n} a^{r} b^{n-r} = \sum_{r=0}^{n} 2^{r} (\frac{1}{2})^{n-r} = \sum_{r=0}^{n} 2^{r} \cdot 2^{r-n} = \sum_{r=0}^{n} 2^{2r-n} = 2^{-n} \sum_{r=0}^{n} (2^{2})^{r} = 2^{-n} \sum_{r=0}^{n} 4^{r}$. This is a geometric series with $n+1$ terms, first term $1$, and common ratio $4$. The sum is $2^{-n} \left[ \frac{1(4^{n+1} - 1)}{4 - 1} ight] = 2^{-n} \left[ \frac{4^{n+1} - 1}{3} ight] = \frac{4^{n+1} - 1}{3 \cdot 2^{n}}$.

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

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