Q54 / 170 · Manipal (MET) 2018 · Mathematics
Let $U_n=2+2^3+2^5+\ldots .+2^{2 n+1}$ and $V_n=1+4+4^2+\ldots .+4^{n-1}$. Then, $\lim _{n \rightarrow \infty} \frac{U_n}{V_n}$ is equal to
- Correct answerA8
- B6
- C$\frac{3}{4}$
- D$\frac{3}{2}$
Solution
Given, $U_n=2+2^3+2^5+2^7+\ldots+2^{2 n+1}$
and $V_n=1+4+4^2+\ldots+4^{n-1}$
$\therefore \quad U_n=\frac{l r-a}{r-1}=\frac{2^{2 n+1} \cdot 4-2}{4-1}=\frac{2^{2 n} \cdot 8-2}{3}$
and $\quad ' V_n=\frac{l r-a}{r-1}=\frac{4^{n-1} \cdot 4-1}{4-1}=\frac{2^{2 n}-1}{3}$
Now, $\quad \frac{U_n}{V_n}=\frac{2^{2 n} \cdot 8-2}{2^{2 n}-1}=\frac{8-\frac{2}{2^{2 n}}}{1-\frac{1}{2^{2 n}}}$
$\therefore \quad \lim _{n \rightarrow \infty} \frac{U_n}{V_n}=\lim _{n \rightarrow \infty} \frac{8-\frac{2}{2^{2 n}}}{1-\frac{1}{2^{2 n}}}=\frac{8-0}{1-0}=8$
Hence, the correct answer is option: 'A'.