What is maclaurin (1+x)^{-4/5} ?

The answer to maclaurin (1+x)^{-4/5} is 1-4/5 x+18/25 x^2-84/125 x^3+399/625 x^4+\ldots

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maclaurin (1+x)^{-4/5}

maclaurin

(1 + x)^{- \frac{4}{5}}

1 - \frac{4}{5} x + \frac{18}{25} x^{2} - \frac{84}{125} x^{3} + \frac{399}{625} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{- \frac{4}{5}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{- \frac{4}{5}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{- \frac{4}{5}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- \frac{4}{5}}{1 !} x + \frac{\frac{36}{25}}{2 !} x^{2} + \frac{- \frac{504}{125}}{3 !} x^{3} + \frac{\frac{9576}{625}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{4}{5} x + \frac{18}{25} x^{2} - \frac{84}{125} x^{3} + \frac{399}{625} x^{4} + \dots

\frac{\partial}{\partial y} (e^{- \frac{y}{x}})

\frac{\partial}{\partial y} (x^{- 2} y^{- 3})

\frac{1}{2} x 25 - x^{2}

\int (\frac{3 x ^{3} - 4 x - 2}{x ^{3}}) d x

f (x) = \frac{6 x}{x ^{2} + 1}, (1, 3)

What is maclaurin (1+x)^{-4/5} ?
The answer to maclaurin (1+x)^{-4/5} is 1-4/5 x+18/25 x^2-84/125 x^3+399/625 x^4+\ldots