What is maclaurin 1/((1+x)^2) ?

The answer to maclaurin 1/((1+x)^2) is 1-2x+3x^2-4x^3+5x^4+\ldots

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maclaurin 1/((1+x)^2)

maclaurin

\frac{1}{( 1 + x ) ^{2}}

1 - 2 x + 3 x^{2} - 4 x^{3} + 5 x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} x + \frac{6}{2 !} x^{2} + \frac{- 24}{3 !} x^{3} + \frac{120}{4 !} x^{4} + \dots

Refine

= 1 - 2 x + 3 x^{2} - 4 x^{3} + 5 x^{4} + \dots

\frac{\partial}{\partial x} (14 x)

\frac{d}{d x} (e^{2 x - 1} ln (2 x - 1))

\int \frac{9}{( x ^{2} - 1 ) ^{2}} d x

sec^{7} (x)

4 - x^{2}

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