What is maclaurin (1+x)^{-2} ?

The answer to maclaurin (1+x)^{-2} is 1-2x+3x^2-4x^3+5x^4+\ldots

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maclaurin (1+x)^{-2}

maclaurin

(1 + x)^{- 2}

1 - 2 x + 3 x^{2} - 4 x^{3} + 5 x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} x + \frac{6}{2 !} x^{2} + \frac{- 24}{3 !} x^{3} + \frac{120}{4 !} x^{4} + \dots

Refine

= 1 - 2 x + 3 x^{2} - 4 x^{3} + 5 x^{4} + \dots

n \to \infty lim (\frac{2 n ^{5} - 3 n ^{4} + 8}{3 n ^{5} + 1})

f (x) = 3 x - 4

\int_{1}^{2} 6 e^{6 x} d x

(2 x)^{6 x}

\int 12 cot^{4} (x) d x

What is maclaurin (1+x)^{-2} ?
The answer to maclaurin (1+x)^{-2} is 1-2x+3x^2-4x^3+5x^4+\ldots