What is taylor 1/((1-x)^2),0 ?

The answer to taylor 1/((1-x)^2),0 is 1+2x+3x^2+4x^3+5x^4+\ldots

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taylor 1/((1-x)^2),0

taylor

\frac{1}{( 1 - x ) ^{2}}, 0

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

h (t) = - 16 t^{2} + 144 t

30 e^{- 0.0000323 h}

x \to 0 lim (- 2 x sin (\frac{1}{x}))

\frac{\partial}{\partial y} (x y^{2} - 6 x^{2} - 3 y^{2})

y = (8 x^{3} + 5)^{\frac{3}{2}}

What is taylor 1/((1-x)^2),0 ?
The answer to taylor 1/((1-x)^2),0 is 1+2x+3x^2+4x^3+5x^4+\ldots