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

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

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taylor 1/(2-x),1

taylor

\frac{1}{2 - x}, 1

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \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

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

\frac{d}{d x} (\frac{3 x ^{3} + x + 2}{5 x ^{2} + 1})

\frac{\partial}{\partial x} (7 ln (x^{2} + y^{2}))

n = 5 \sum \infty \frac{1}{n}

x^{^{''}} - 5 x^{^{'}} + 6 x = e^{t}

\frac{d}{d x} (- \frac{7}{x ^{8}})

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