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

\int (4 x^{2} + 3) d x

\int_{- 2}^{1} \frac{x ^{2} + 4 x + 8}{x - 2} d x

g (x) = cos (x) + 1, f (x) = sin (x)

\int_{1}^{9} 3 - x d x

f (x) = x^{3} (x - 5)^{2}

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