Question

If $a=\min \left\{x^{2}+4 x+5, x \in R\right\}$ and $b=\displaystyle\lim _{\theta \rightarrow 0} \frac{1-\cos 2 \theta}{\theta^{2}}$, then the value of $\displaystyle\sum_{r=0}^{n} a^{r} \cdot b^{n-r}$ is

• A. $\frac{2^{n+1}-1}{4 \cdot 2^{n}}$
• B. $2^{n+1}-1$
• C. $\frac{2^{n+1}-1}{3 \cdot 2^{n}}$
• D. None of these

Answer 1

Answer: (B)

$x^{2}+4 x+5=(x+2)^{2}+1 \geq 1 . \text { So, } a=1$
Also, $b=\displaystyle\lim _{\theta \rightarrow 0} \frac{1-\cos 2 \theta}{\theta^{2}}$
$=\displaystyle\lim _{\theta \rightarrow 0} \frac{2 \sin ^{2} \theta}{\theta^{2}}=2$
$\therefore \displaystyle\sum_{r=0}^{n} a^{r} \cdot b^{n-r}=b^{n}+a b^{n-1}+a^{2} b^{n-2}+\ldots+a^{n}$
$=\frac{b^{n}\left[1-\left(\frac{a}{b}\right)^{n+1}\right]}{1-\frac{a}{b}}$
$=\frac{2^{n}\left[1-\left(\frac{1}{2}\right)^{n+1}\right]}{1-\frac{1}{2}}$
$=\frac{2^{n+1}\left(2^{n+1}-1\right)}{2^{n+1}}$
$=\left(2^{n+1}-1\right)$