Question

Consider the function y=\ ln(1+x).

Which of the following is the Maclaurin series expansion of the first three terms?

1. $x-\frac{x^2}{2}+\frac{x^3}{6}$
2. $x-\frac{x^2}{3}+\frac{x^3}{2}$
3. $x-\frac{x^2}{2}+\frac{x^3}{3}​$
4. $x+\frac{x^2}{2}-\frac{x^3}{3}$

Answer 1

We should obtain the first three terms of the series ${\displaystyle f(x)=\sum _{n=0}^{+\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}}$

$f^{(0)}(0) = \ln (1+0) = 0, \\ f^{(1)}(0) = \frac{1}{1+0} = 1, \\ f^{(2)}(0) = -\frac{1}{(1+0)^2} = -1, \\ f^{(3)}(0) = \frac{2}{(1+0)^3} = 2$

$\ln(1+x) = 0 + x - \frac{1}{2}x^2 + \frac{2}{6}x^3 = x - \frac{1}{2}x^2 + \frac{1}{3}x^3.$ The correct answer is 3