What is maclaurin f(x)= 1/((1-x)^2) ?

The answer to maclaurin f(x)= 1/((1-x)^2) is 1+2x+3x^2+4x^3+5x^4+\ldots

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maclaurin f(x)= 1/((1-x)^2)

maclaurin

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

x \to 0 + lim (2 (tan (x))^{x})

\frac{d y}{d x} = \frac{( x ^{2} + y ^{2} )}{2 x y}

y^{^{''}} - y = \frac{e ^{x}}{e ^{x} + 1}

\frac{d}{d x} (ln (\frac{x + 1}{x - 2}))

\int_{1}^{\infty} \frac{5}{x ^{4}} d x

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