Question

Find the first four non-zero terms for the Maclaurin series of the function [f(x) = ln(1 + 2x)].

A) (2x - 2x^2 + frac{8}{3}x^3 - frac{16}{4}x^4)
B) (2x + 2x^2 + 2x^3 + 2x^4)
C) (2x - 2x^2 + 2x^3 - 2x^4)
D) (2x + 2x^2 - 2x^3 - 2x^4)

Answer 1

The Maclaurin series for the function
`\(f(x) = \ln(1 + 2x)\)` is accurately represented by the expression
`\(2x - 2x^2 + (8)/(3)x^3 - (16)/(4)x^4\),` making option A the correct choice. This result is obtained by substituting
`\(2x\)` into the Taylor series expansion formula for the natural logarithm and simplifying the terms to capture the first four non-zero terms.

Step-by-step explanation:

Certainly! Let's find the Maclaurin series for the function
`\(f(x) = \ln(1 + 2x)\)`in detail.

The Maclaurin series for
`\(\ln(1 + x)\)` is given by the formula:

`\[ \ln(1 + x) = x - (x^2)/(2) + (x^3)/(3) - (x^4)/(4) + \ldots \]`

Now, substitute
`\(2x\) for \(x\)` in this formula:

`\[ f(x) = \ln(1 + 2x) = 2x - ((2x)^2)/(2) + ((2x)^3)/(3) - ((2x)^4)/(4) + \ldots \]`

Simplify each term:

`\[ = 2x - 2x^2 + (8)/(3)x^3 - (16)/(4)x^4 + \ldots \]`

Therefore, the first four non-zero terms of the Maclaurin series for
`\(f(x) = \ln(1 + 2x)\) are \(2x - 2x^2 + (8)/(3)x^3 - (16)/(4)x^4\).`

Understanding the pattern and coefficients in the Taylor series expansion of
`\(\ln(1 + x)\)` allows us to confidently substitute
`\(2x\)` into the formula to find the Maclaurin series for
`\(f(x) = \ln(1 + 2x)\)` . The calculation involves basic algebraic operations and the application of the general formula for the Maclaurin series of the natural logarithm function.

So correct option is option A