If |x|

Question

(C) The expansion of $\ln(1 - x)$ for $|x| < 1$ is given by $\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots$N

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(C) The expansion of $\ln(1 - x)$ for $|x| Now,consider the expression $(1 - x) \ln(1 - x)$:$(1 - x) \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots ight)$To find the coefficient of $x^5$,we multiply the terms:$1 	imes (	ext{coefficient of } x^5 	ext{ in } \ln(1 - x)) - x 	imes (	ext{coefficient of } x^4 	ext{ in } \ln(1 - x))$$= 1 	imes \left( -\frac{1}{5} ight) - 1 	imes \left( -\frac{1}{4} ight)$$= -\frac{1}{5} + \frac{1}{4}$$= \frac{-4 + 5}{20} = \frac{1}{20}$$= 0.05$

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $(1 - x) \ln(1 - x)$ is

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