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

The answer to maclaurin (1+x)^{-1} is 1-x+x^2-x^3+x^4+\ldots

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maclaurin (1+x)^{-1}

maclaurin

(1 + x)^{- 1}

1 - x + x^{2} - x^{3} + x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 1}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \dots

Refine

= 1 - x + x^{2} - x^{3} + x^{4} + \dots

\int \frac{2 x}{x ^{2} - 16} + \frac{1}{x} + 2 d x

y^{^{'}} = x

f (x) = (- 2 x^{2} + 4)^{4}

y = (\frac{( x + 1 )}{( x + 3 ) ^{2}})

\frac{d y}{d t} = e^{t - y}

What is maclaurin (1+x)^{-1} ?
The answer to maclaurin (1+x)^{-1} is 1-x+x^2-x^3+x^4+\ldots