Q103 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
$\begin{aligned} & \lim _{n \rightarrow \infty} n^{-n k} \\ & \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n=\end{aligned}$
- A2
- Correct answerB$e^{2\left(1-\frac{1}{2^k}\right)}$
- C$2\left(1-\frac{1}{2^k}\right)$
- D$e^2$
Solution
$\begin{aligned} & \lim _{n \rightarrow \infty} n^{-n k}\left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n \\ & =P \text { Let } \\ & \Rightarrow P=\lim _{n \rightarrow \infty}\left(\frac{\left.(n+1)\left(n+\frac{1}{2^1}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right)^n}{n^k}\right)\end{aligned}$
$$
=\lim _{n \rightarrow \infty}\left[\left(\frac{n+\frac{1}{2^0}}{n}\right)\left(\frac{n+\frac{1}{2^1}}{n}\right)\left(\frac{n+\frac{1}{2^2}}{n}\right) \ldots\left(\frac{n+\frac{1}{2^{k-1}}}{n}\right)\right]^n
$$
So,
$$
\begin{aligned}
& \log P=\lim _{n \rightarrow \infty} n\left[\begin{array}{l}
\log \left(1+\frac{1}{2^0 n}\right)+\log \left(1+\frac{1}{2^1 n}\right) \\
+\log \left(1+\frac{1}{2^2 n}\right)+\ldots+\log \left(1+\frac{1}{2^{k-1} n}\right)
\end{array}\right] \\
& =\lim _{n \rightarrow \infty}\left[\begin{array}{c}
\left.\frac{\log \left(1+\frac{1}{2^{\circ} n}\right)}{1 / n}+\frac{\log \left(1+\frac{1}{2^1 n}\right)}{1 / n}+\frac{\log \left(1+\frac{1}{2^2 n}\right)}{1 / n}\right] \\
+\ldots+\frac{\log \left(1+\frac{1}{2^{k-1} n}\right)}{1 / n}
\end{array}\right] \\
& =\frac{1}{2^{\circ}}+\frac{1}{2^1}+\frac{1}{2^2}+\ldots .+\frac{1}{2^{k-1}}=\frac{1-\frac{1}{2^k}}{1-\frac{1}{2}}=2\left(1-\frac{1}{2^k}\right)
\end{aligned}
$$
So, $\quad P=e^{2\left(1-\frac{1}{2^k}\right)}$
Hence, the correct answer is option: 'B'.