Q72 / 120 · VITEEE 2012 · Mathematics
The coefficient of $\mathrm{x}^{\mathrm{n}}$ in the expansion of $\log _{a}(1+x)$ is
- A$\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}}$
- Correct answerB$\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}$
- C$\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{e}} \mathrm{a}$
- D$\frac{(-1)^{\mathrm{n}}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}$
Solution
$$
\begin{array}{l}
\log _{\mathrm{a}}(1+\mathrm{x})=\log _{\mathrm{e}}(1+\mathrm{x}) \log _{\mathrm{a}} \mathrm{e} \\
=\log _{\mathrm{a}} \mathrm{e}\left[\sum_{\mathrm{n}=1}^{\infty}(-1)^{\mathrm{n}-1} \frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n}}\right]
\end{array}
$$
So, the coefficient of $x^{n}$ in $\log _{a}(1+x)$ is
$$
\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}
$$
Hence, the correct answer is option: 'B'.