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

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(D) $a = (x+1)^2 + 2$. Since $(x+1)^2 \ge 0$,the minimum value $a = 2$. $b = \lim_{x 	o 0} \frac{\sin x \cos x}{e^x - e^{-x}} = \lim_{x 	o 0} \frac{

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

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(D) $a = (x+1)^2 + 2$. Since $(x+1)^2 \ge 0$,the minimum value $a = 2$. $b = \lim_{x 	o 0} \frac{\sin x \cos x}{e^x - e^{-x}} = \lim_{x 	o 0} \frac{\frac{1}{2} \sin 2x}{2 \sinh x} = \lim_{x 	o 0} \frac{\sin 2x}{4 \sinh x} = \frac{1}{4} \lim_{x 	o 0} \frac{2x}{x} = \frac{1}{2}$. Now,$\sum_{r=0}^n a^r b^{n-r} = \sum_{r=0}^n 2^r (\frac{1}{2})^{n-r} = \frac{1}{2^n} \sum_{r=0}^n 2^r 2^r = \frac{1}{2^n} \sum_{r=0}^n 4^r$. This is a geometric progression with $n+1$ terms,first term $1$,and common ratio $4$. Sum $= \frac{1}{2^n} \left( \frac{4^{n+1} - 1}{4 - 1} ight) = \frac{4^{n+1} - 1}{3 \cdot 2^n}$.

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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