What is maclaurin 1/(1-x),3 ?

The answer to maclaurin 1/(1-x),3 is sum from n=0 to infinity of x^n

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maclaurin 1/(1-x),3

maclaurin

\frac{1}{1 - x}, 3

n = 0 \sum \infty x^{n}

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

The series sum representation

= n = 0 \sum \infty x^{n}

\int_{0}^{1} \frac{2}{x - 2} d x

\frac{d}{d x} (\frac{1}{1 + 2 x})

\frac{d}{d x} (9 + 5 x)

\frac{d}{d x} (2 x^{- 7 3 x})

n = 1 \sum \infty n! 2^{n} n^{- n}

What is maclaurin 1/(1-x),3 ?
The answer to maclaurin 1/(1-x),3 is sum from n=0 to infinity of x^n