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

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(D) First,find $a$: $a = \min \{x^2 + 2x + 3\}$. Since $x^2 + 2x + 3 = (x+1)^2 + 2$,the minimum value is $a = 2$.Next,find $b$: $b = \lim_{x 	o 0} \f

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

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(D) First,find $a$: $a = \min \{x^2 + 2x + 3\}$. Since $x^2 + 2x + 3 = (x+1)^2 + 2$,the minimum value is $a = 2$.Next,find $b$: $b = \lim_{x 	o 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Using $L$'Hopital's rule or standard limits,$b = \lim_{x 	o 0} \frac{\sin x}{x} \cdot \frac{x}{e^x - e^{-x}} \cdot \cos x = 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}$.Now,calculate the sum $S = \sum_{r=0}^n a^r b^{n-r} = b^n + a b^{n-1} + a^2 b^{n-2} + \dots + a^n$. This is a geometric progression with $n+1$ terms,first term $T_1 = b^n$,and common ratio $r = \frac{a}{b} = \frac{2}{1/2} = 4$.The sum is $S = b^n \frac{(a/b)^{n+1} - 1}{(a/b) - 1} = (1/2)^n \frac{4^{n+1} - 1}{4 - 1} = \frac{1}{2^n} \cdot \frac{4^{n+1} - 1}{3} = \frac{4^{n+1} - 1}{3 \cdot 2^n}$.

Let $a = \min \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Then the value of $\sum_{r=0}^n a^r b^{n-r}$ is

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