What is maclaurin 1/(1+x+x^2) ?

The answer to maclaurin 1/(1+x+x^2) is 1-x+x^3-x^4+x^6+\ldots

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maclaurin 1/(1+x+x^2)

maclaurin

\frac{1}{1 + x + x ^{2}}

1 - x + x^{3} - x^{4} + x^{6} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{- 24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{720}{6 !} x^{6} + \dots

Refine

= 1 - x + x^{3} - x^{4} + x^{6} + \dots

12 x^{5} - 8 x^{4} + 20 x^{3}

(- 1, 8), (3, - 4)

f (x) = (3 x^{3} - 4) (- 4 x - 3)

y = cos (x), y = 0.5

\int_{- 10}^{10} x e^{- x^{8}} d x

What is maclaurin 1/(1+x+x^2) ?
The answer to maclaurin 1/(1+x+x^2) is 1-x+x^3-x^4+x^6+\ldots