What is maclaurin 1/(1+3x) ?

The answer to maclaurin 1/(1+3x) is 1-3x+9x^2-27x^3+81x^4+\ldots

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maclaurin 1/(1+3x)

maclaurin

\frac{1}{1 + 3 x}

1 - 3 x + 9 x^{2} - 27 x^{3} + 81 x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 3}{1 !} x + \frac{18}{2 !} x^{2} + \frac{- 162}{3 !} x^{3} + \frac{1944}{4 !} x^{4} + \dots

Refine

= 1 - 3 x + 9 x^{2} - 27 x^{3} + 81 x^{4} + \dots

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

ln (x)

2 y y^{^{'}} + 6 = y^{2} + 6 x

\int_{3}^{4} (5 - x^{2}) - (17 - 7 x) d x

\frac{\partial}{\partial x} (x^{2} tan (x y))

What is maclaurin 1/(1+3x) ?
The answer to maclaurin 1/(1+3x) is 1-3x+9x^2-27x^3+81x^4+\ldots