What is maclaurin 1/(2-x) ?

The answer to maclaurin 1/(2-x) is 1/2+1/4 x+1/8 x^2+1/16 x^3+1/32 x^4+\ldots

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maclaurin 1/(2-x)

maclaurin

\frac{1}{2 - x}

\frac{1}{2} + \frac{1}{4} x + \frac{1}{8} x^{2} + \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= \frac{1}{2} + \frac{1}{4} x + \frac{1}{8} x^{2} + \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

g (x) = \frac{- 7}{x}

x \to - 2 lim (3 x^{3} - 2 x + 7)

\int \frac{1}{t - 20} d t

\int \frac{1}{1 + e ^{- 5 x}} d x

\frac{d}{d x} (sin^{2 x} (x))

What is maclaurin 1/(2-x) ?
The answer to maclaurin 1/(2-x) is 1/2+1/4 x+1/8 x^2+1/16 x^3+1/32 x^4+\ldots