What is maclaurin ln(1-2y) ?

The answer to maclaurin ln(1-2y) is -2y-2y^2-8/3 y^3-4y^4-32/5 y^5+\ldots

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maclaurin ln(1-2y)

maclaurin

ln (1 - 2 y)

- 2 y - 2 y^{2} - \frac{8}{3} y^{3} - 4 y^{4} - \frac{32}{5} y^{5} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d y} ( ln ( 1 - 2 y ) ) ( 0 )}{1 !} y + \frac{\frac{d ^{2}}{d y ^{2}} ( ln ( 1 - 2 y ) ) ( 0 )}{2 !} y^{2} + \frac{\frac{d ^{3}}{d y ^{3}} ( ln ( 1 - 2 y ) ) ( 0 )}{3 !} y^{3} + \dots

Evaluate Derivatives

= 0 + \frac{- 2}{1 !} y + \frac{- 4}{2 !} y^{2} + \frac{- 16}{3 !} y^{3} + \frac{- 96}{4 !} y^{4} + \frac{- 768}{5 !} y^{5} + \dots

Refine

= - 2 y - 2 y^{2} - \frac{8}{3} y^{3} - 4 y^{4} - \frac{32}{5} y^{5} + \dots

\int \frac{3 x - 1}{x ^{2} - x - 6} d x

\frac{d}{d x} (2 lo g_{10} (x))

\int_{1}^{2} (4 x^{\frac{3}{2}}) d x

\frac{\partial}{\partial y} (sin (2 x + 5 y))

What is maclaurin ln(1-2y) ?
The answer to maclaurin ln(1-2y) is -2y-2y^2-8/3 y^3-4y^4-32/5 y^5+\ldots