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

The answer to maclaurin (1+x)^{-3} is 1-3x+6x^2-10x^3+15x^4+\ldots

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maclaurin (1+x)^{-3}

maclaurin

(1 + x)^{- 3}

1 - 3 x + 6 x^{2} - 10 x^{3} + 15 x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 3}{1 !} x + \frac{12}{2 !} x^{2} + \frac{- 60}{3 !} x^{3} + \frac{360}{4 !} x^{4} + \dots

Refine

= 1 - 3 x + 6 x^{2} - 10 x^{3} + 15 x^{4} + \dots

\int - cos (x) sin (2 x) d x

\int 5 1 - x^{2} d x

\frac{\partial}{\partial x} (3 (x - 1) 2 + 2 (y + 3) 2 + 7)

\int (e^{- 5 x}) d x

x y^{2} \frac{d y}{d x} = x^{3} + y^{3}

What is maclaurin (1+x)^{-3} ?
The answer to maclaurin (1+x)^{-3} is 1-3x+6x^2-10x^3+15x^4+\ldots