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

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

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

taylor

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

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

Solution steps

Apply the Taylor 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

\int x^{5} cos (x^{2}) d x

x \to 1 - lim (\frac{7}{1 - x})

\int \frac{x}{1 - 4 x ^{2}} d x

\int \frac{5 x}{( 7 - x ^{2} ) ^{3}} d x

n = 1 \sum \infty \frac{1}{n ^{1.5}}

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