Q91 / 155 · WBJEE 2019 · Mathematics
Let $a=\min \left\{x^{2}+2 x+3: x \in R\right\}$ and $b=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}} \cdot$ Then $\sum_{r=0}^{n} a^{r} b^{n-r}$ is
- A$\frac{2^{n+1}-1}{3 \cdot 2^{n}}$
- B$\frac{2^{n+1}+1}{3 \cdot 2^{n}}$
- Correct answerC$\frac{4^{n+1}-1}{3 \cdot 2^{n}}$
- D$\frac{1}{2}\left(2^{n}-1\right)$
Solution
$$
\begin{aligned}
&\text { (c) Let } f(x)=x^{2}+2 x+3\\
&\begin{aligned}
a &=f(x)_{\min }=\frac{-D}{4 a}=\frac{-(4-12}{4}=\frac{8}{4}=2 \\
\text { and } b &=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}}=\lim _{\theta \rightarrow 0} \frac{1-1+2 \sin ^{2} \theta / 2}{\theta^{2}} \\
&=\lim _{\theta \rightarrow 0} \frac{2 \sin ^{2} \theta / 2}{(\theta / 2)^{2} \cdot 4}=\lim _{\theta \rightarrow 0} \frac{1}{2} \cdot\left[\frac{\sin ^{2} \theta / 2}{(\theta / 2)^{2}}\right] \\
&=\frac{1}{2} \cdot \lim _{\theta \rightarrow 0} \frac{\sin ^{2} \theta / 2}{(\theta / 2)^{2}} \\
&=\frac{1}{2} \cdot 1=\frac{1}{2} \\
b &=\frac{1}{2} \\
\text { Now, } & \sum_{r=0}^{n} a^{r} \cdot b^{n-r}
\end{aligned}
\end{aligned}
$$
$=\sum_{r=0}^{n}\left(2^{r}\left(\frac{1}{2}\right)^{n-r}\right)$
$=\sum_{r=0}^{n} 2^{\prime} \cdot 2^{(r-n)}$
$=\sum_{r=0}^{n} 2^{2 r-n}$
$=2^{-n} \sum_{r=0}^{n} 2^{2 r}$
$=2^{-n}\left[1+2^{2}+2^{4}+2^{6}+\ldots+2^{2 n}\right]$
$=2^{-n}\left[\frac{1 \cdot\left(2^{2}\right)^{n+1}-1}{2^{2}-1}\right]\left\{\because S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\right\}$
$=2^{-n}\left[\frac{4^{n+1}-1}{3}\right]$
$=\frac{4^{n+1}-1}{3 \cdot 2^{n}}$
Hence, the correct answer is option: 'C'.