Q175 / 270 · NDA 2017 (Phase 2) · Mathematics
The value of $\frac{1}{\log _3 e}+\frac{1}{\log _3 e^2}+\frac{1}{\log _3 e^4}+\ldots$. up to infinite terms is
- Correct answerA$\log _e 9$
- B$0$
- C$1$
- D$\log _e 3$
Solution
$\frac{1}{\log _3 \mathrm{e}}+\frac{1}{\log _3 \mathrm{e}^2}+\frac{1}{\log _3 \mathrm{e}^4}+\ldots$
$=\frac{1}{\log _3 \mathrm{e}}+\frac{1}{2 \log _3 \mathrm{e}}+\frac{1}{4 \log _3 \mathrm{e}}+\ldots \cdot\left(\right.$ Since, $\left.\log _{\mathrm{a}} \mathrm{b}^{\mathrm{m}}=\mathrm{m} \mathrm{log}_{\mathrm{a}} \mathrm{b}\right)$
$=\log _{\mathrm{e}} 3+\frac{\log _{\mathrm{e}} 3}{2}+\frac{\log _{\mathrm{e}} 3}{4}+\ldots . \quad\left(\right.$ Since, $\left.\log _{\mathrm{a}} \mathrm{b}=\frac{1}{\log _{\mathrm{b}} \mathrm{a}}\right)$
$=\log _{\mathrm{e}} 3\left(1+\frac{1}{2}+\frac{1}{4}+\ldots.\right)$
$=\log _{\mathrm{e}} 3\left(\frac{1}{1-\frac{1}{2}}\right)\left(\because 1, \frac{1}{2}, \frac{1}{4} \ldots\right.$ is G.P. with $\left.\mathrm{a}=1, \mathrm{r}=\frac{1}{2}\right)$
$=\log _{\mathrm{e}} 3(2)=2 \log _{\mathrm{e}} 3=\log _{\mathrm{e}} 3^2=\log _{\mathrm{e}} 9$.
Hence, the correct answer is option: 'A'.