Using the first four terms of the Maclaurin series of $\ln ( 1 + x ),$ the approximated value of ln 2 is

A) $\frac { 7 } { 12 }$
B) $\frac { 7 } { 16 }$
C) $\frac { 7 } { 3 }$
D) $\frac { 2 } { 3 }$
E) $\frac { 1 } { 3 }$

Question

Using the first four terms of the Maclaurin series of $\ln ( 1 + x ),$ the approximated value of ln 2 is

A) $\frac { 7 } { 12 }$
B) $\frac { 7 } { 16 }$
C) $\frac { 7 } { 3 }$
D) $\frac { 2 } { 3 }$
E) $\frac { 1 } { 3 }$

Quizplus · Accepted Answer

The Maclaurin series for $\ln(1+x)$ is $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$. Using the first four terms to approximate $\ln(2)$, we set $x = 1$, giving $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{2}{3} - \frac{3}{4} = \frac{8 - 9}{12} = \frac{-1}{12}$. However, the correct calculation should be $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$, indicating a mistake in my initial subtraction. Correctly, it's $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{12}{12} - \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12}$.

Using the First Four Terms of the Maclaurin Series Of \ln