What is maclaurin (1+x)^{-5/3} ?

The answer to maclaurin (1+x)^{-5/3} is 1-5/3 x+20/9 x^2-220/81 x^3+770/243 x^4+\ldots

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maclaurin (1+x)^{-5/3}

maclaurin

(1 + x)^{- \frac{5}{3}}

1 - \frac{5}{3} x + \frac{20}{9} x^{2} - \frac{220}{81} x^{3} + \frac{770}{243} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{5}{3}}{1 !} x + \frac{\frac{40}{9}}{2 !} x^{2} + \frac{- \frac{440}{27}}{3 !} x^{3} + \frac{\frac{6160}{81}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{5}{3} x + \frac{20}{9} x^{2} - \frac{220}{81} x^{3} + \frac{770}{243} x^{4} + \dots

\frac{\partial}{\partial x} (ln (x y))

\frac{d}{d x} (x^{3} e^{- x})

x \to 9 lim (x + 7)

\frac{d}{d t} (1 0^{2 t})

\frac{d}{d x} (3^{t} (5 t^{4} + 7 t + y (t, x) (t, x)^{4}))

What is maclaurin (1+x)^{-5/3} ?
The answer to maclaurin (1+x)^{-5/3} is 1-5/3 x+20/9 x^2-220/81 x^3+770/243 x^4+\ldots