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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
$\frac{{{2^{n + 1}} - 1}}{{{{3.2}^n}}}$
$\frac{{{2^{n + 1}} + 1}}{{{{3.2}^n}}}$
$\frac{{{4^{n + 1}} - 1}}{{{{3.2}^n}}}$
None of these
Solution
Let $a=\min \left\{x^{2}+2 x+3: x \in R\right\}$ and $b=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}}$
Now, $x^{2}+2 x+3=x^{2}+2 x+1+2=(x+1)^{2}+2$
$\therefore \min \left\{x^{2}+2 x+3: x \in R\right\}=\min \left\{(x+1)^{2}+2: x \in R\right\}=2$
Also, $b=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \theta}{2 \theta}=\frac{1}{2}=\left\{\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right\}$
$\sum_{r=0}^{n} a^{r} \cdot b^{n-r}$
$=\sum_{r=0}^{n} 2^{r}\left(\frac{1}{2}\right)^{n-r}$
$=\quad 2^{-n} \sum_{r=0}^{n} 2^{2 r}$
$=\quad 2^{-n}\left\{1+2^{2}+2^{4}+\ldots . .2^{2 n}\right\}$
$=\frac{4^{n+1}-1}{3.2^{n}}$
