What is taylor 1/(1-\frac{1){1-x}} ?

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

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taylor 1/(1-\frac{1){1-x}}

taylor

\frac{1}{1 - \frac{1}{1 - x}}

2 + (x + 1) + (x + 1)^{2} + (x + 1)^{3} + (x + 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= 2 + (x + 1) + (x + 1)^{2} + (x + 1)^{3} + (x + 1)^{4} + \dots

\frac{\partial}{\partial y} (tan (5 + 2 x^{2} y^{4} z^{2}))

\frac{\partial}{\partial y} (e^{y} x)

\int cos (θ) sin^{3} (θ) d θ

9 y^{^{''}} + 32 y = 0, y^{^{'}} (0) = 9, y (0) = 5

(x - 8 y) d x = - 4 x d y

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