What is maclaurin ln(1+x),\at (x=2) ?

The answer to maclaurin ln(1+x),\at (x=2) is sum from n=1 to infinity}(-1)^{n+1 of (x^n)/n

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maclaurin ln(1+x),\at (x=2)

maclaurin

ln (1 + x), at (x = 2)

n = 1 \sum \infty (- 1)^{n + 1} \frac{x ^{n}}{n}

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

The series sum representation

= n = 1 \sum \infty (- 1)^{n + 1} \frac{x ^{n}}{n}

3 x^{2} - 4 x

e^{x - 1}, 0

y + sin (x) = x^{3} y^{^{'}}

\int e^{4 y} d y

x \to 0 + lim (x^{6 x})

What is maclaurin ln(1+x),\at (x=2) ?
The answer to maclaurin ln(1+x),\at (x=2) is sum from n=1 to infinity}(-1)^{n+1 of (x^n)/n