Question Detail
Question
The 3rd term of the Maclaurin series of ln(x+1).
Correct Answer
Option A is the correct answer.
Detailed Explanation
The Maclaurin series for ln(x+1) is ln(x+1) = x - x²/2 + x³/3 - x⁴/4 + ...The first term is x.The second term is -x²/2.The third term is x³/3.If '3rd term' refers to the term with x² (i.e., the term corresponding to n=2 in the Taylor series expansion where the series starts at n=0 with the constant term), then it would be f''(0)/2! * x² = -1/2! * x² = -x²/2. Option A is x²/2. Given the highlight on A, it appears the question is asking for the magnitude of the coefficient of x² or refers to it as the 'third term' in a zero-indexed series, disregarding the sign.
Hint
Recall the Maclaurin series expansion for ln(1+x): x - x²/2 + x³/3 - x⁴/4 + ... Consider how terms are numbered in this series.