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

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Question

Let $a=\min \left\{x^{2}+2 x+3: x \in Right\}$ and $b=\lim _{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^{2}}$

Now, $x^{2}+2 x+3=x^{2}+2 x+

Vedclass · Accepted Answer

$\frac{{{4^{n + 1}} - 1}}{{{{3.2}^n}}}$

Vedclass · Answer

Let $a=\min \left\{x^{2}+2 x+3: x \in Right\}$ and $b=\lim _{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^{2}}$

Now, $x^{2}+2 x+3=x^{2}+2 x+1+2=(x+1)^{2}+2$

$	herefore \min \left\{x^{2}+2 x+3: x \in Right\}=\min \left\{(x+1)^{2}+2: x \in Right\}=2$

Also, $b=\lim _{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^{2}}$

$=\lim _{	heta ightarrow 0} \frac{\sin 	heta}{2 	heta}=\frac{1}{2}=\left\{\lim _{	heta ightarrow 0} \frac{\sin 	heta}{	heta}=1ight\}$

$\sum_{r=0}^{n} a^{r} \cdot b^{n-r}$

$=\sum_{r=0}^{n} 2^{r}\left(\frac{1}{2}ight)^{n-r}$

$=\quad 2^{-n} \sum_{r=0}^{n} 2^{2 r}$

$=\quad 2^{-n}\left\{1+2^{2}+2^{4}+\ldots . .2^{2 n}ight\}$

$=\frac{4^{n+1}-1}{3.2^{n}}$

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

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