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Question #a8879

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Apr 6, 2017

A formula to memorize is the MacLaurin series of $ln (1 + x)$ $ln(1+x)$ .

$ln (1 + x) = \infty \sum n = 0 \frac{{(- 1)}^{n} {(x)}^{n + 1}}{n + 1}$ $ln(1+x)=sum_(n=0)^(oo)frac{(-1)^n(x)^(n+1)}{n+1}$

Therefore:
$ln (1 + 4 x) = \infty \sum n = 0 \frac{{(- 1)}^{n} {(4 x)}^{n + 1}}{n + 1}$ $ln(1+4x)=sum_(n=0)^oofrac{(-1)^n(4x)^(n+1)}{n+1}$

$= \infty \sum n = 0 \frac{{(- 1)}^{n} {(4)}^{n + 1} {(x)}^{n + 1}}{n + 1}$ $=sum_(n=0)^oofrac{(-1)^n(4)^(n+1)(x)^(n+1)}{n+1}$

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