The coefficient of $x^{n}$ in the expansion of $l o g_{a} (1 + x)$ is

\frac{(- 1)^{n - 1}}{n}

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\frac{(- 1)^{n - 1}}{n} l o g_{a} e

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\frac{(- 1)^{n - 1}}{n} l o g_{e} a

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\frac{(- 1)^{n}}{n} l o g_{a} e

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Solution

The correct option is D $\frac{(- 1)^{n - 1}}{n} l o g_{a} e$
$l o g_{a} (1 + x) = l o g_{e} (1 + x) l o g_{a} e$

$= l o g_{a} e [\infty \sum n = 1 (- 1)^{n - 1} \frac{x^{n}}{n}]$
So, the coefficient of $x^{n}$ in $l o g_{a} (1 + x)$ is $\frac{(- 1)^{n - 1}}{n} l o g_{a} e$ .

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