Let a = text{Minimum} {x^2 + 2x + 3, x in R} | vedclass

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(C) Given $a = \min \{x^2 + 2x + 3 : x \in R\}$.Since $x^2 + 2x + 3 = (x + 1)^2 + 2$,the minimum value is $a = 2$.Given $b = \lim_{	heta 	o 0} \frac

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

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(C) Given $a = \min \{x^2 + 2x + 3 : x \in R\}$.Since $x^2 + 2x + 3 = (x + 1)^2 + 2$,the minimum value is $a = 2$.Given $b = \lim_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2}$.Using the limit formula $\lim_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we get $b = \frac{1}{2}$.Now,we evaluate the sum $\sum_{r=0}^{n} a^r \cdot b^{n-r} = \sum_{r=0}^{n} 2^r \cdot (\frac{1}{2})^{n-r}$.$= \sum_{r=0}^{n} 2^r \cdot 2^{r-n} = \sum_{r=0}^{n} 2^{2r-n} = 2^{-n} \sum_{r=0}^{n} 4^r$.This is a geometric progression with $n+1$ terms,first term $1$,and common ratio $4$.Sum $= 2^{-n} \cdot \frac{4^{n+1} - 1}{4 - 1} = \frac{4^{n+1} - 1}{3 \cdot 2^n}$.

Let $a = \text{Minimum} \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2}$. The value of $\sum_{r = 0}^n a^r \cdot b^{n - r}$ is

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