Question Detail
Question
The \(n^{th}\) term of the Maclaurin series expansion of \(\cos 3x\) is:
Correct Answer
Option D is the correct answer.
Detailed Explanation
The Maclaurin series expansion for (\cos u) is given by:(\cos u = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!} = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \dots)In this question, (u = 3x). Substituting (3x) for (u) in the general formula, the (n^{th}) term is:(\frac{(-1)^n (3x)^{2n}}{(2n)!}).
Hint
Recall the Maclaurin series for \(\cos u\).