What is taylor 1/(1-2x) ?

The answer to taylor 1/(1-2x) is 1+2x+4x^2+8x^3+16x^4+\ldots

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taylor 1/(1-2x)

taylor

\frac{1}{1 - 2 x}

1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{2}{1 !} x + \frac{8}{2 !} x^{2} + \frac{48}{3 !} x^{3} + \frac{384}{4 !} x^{4} + \dots

Refine

= 1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + \dots

\frac{d}{d x} (\frac{x}{2.4 ( c ^{2} + c x )} (c^{2} - x^{2}))

\int \frac{cos ( 2 t )}{( 1 + sin ( 2 t ) ) ^{4}} d t

x \to - 1 lim (\frac{x + 2}{x ^{3} + 1})

\int x x (1 + x^{2} x)^{4} d x

y^{3} - (6 x + 6) + 3 x y^{2} y^{^{'}} = 0

What is taylor 1/(1-2x) ?
The answer to taylor 1/(1-2x) is 1+2x+4x^2+8x^3+16x^4+\ldots