What is maclaurin 1/(1-x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

maclaurin 1/(1-x)

maclaurin

\frac{1}{1 - x}

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_{e}^{e^{2}} \frac{1}{x ln ( x )} d x

f (x) = \frac{e ^{x}}{sin ( 2 x )}

\frac{d}{d x} (\frac{40}{1 + 79 e ^{- x}})

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

\int \frac{2 x - 1}{4 x ^{2} - 1} d x

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