$n$-th Term for Maclaurin Series On a Calculus BC test I had this morning, I had to find the first five terms and the $n$-th term of the following function:
$$ f(x) = x \cos(3x)$$
According to my instructor, I could've manipulated the common MacLaurin series for $f(x) = \cos x$, which I know, to simplify the whole process.
I had no idea how to do such manipulation, so I brutally, painfully, and slowly took derivative after derivative of the original function to get the five terms. However, I have no clue how to get the n-th term or what it is.
 A: We know that $\cos(y) = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}y^{2n}$.
Setting $y=3x$, we get:
$\cos(3x) = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}3^{2n}x^{2n}$
Multiply this by $x$, and we get:
$x\cos(3x) = x\left( \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}3^{2n}x^{2n}\right) = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}3^{2n}}{(2n)!}x^{2n+1}$
A: The standard Maclaurin series for $\cos$ is
$$1-\frac12x^2+\frac1{4!}x^4+\ldots$$
then replace each $x$ by $3x$ to get
$$1-\frac92x^2+\frac{3^4}{4!}x^4+\ldots$$
and multiply through by $x$ to get
$$x-\frac92x^3+\frac{3^4}{4!}x^5+\ldots$$
A: Hint. Write the first few terms of the series for $\cos t$. Substitute $t = 3x$. Multiply term by term by $x$. Then look for the coefficients of the terms you need. 
This method illustrates two of the "manipulations" your instructor had in mind. Others are integration and differentiation term by term, multiplying power series, even inverting them (formally) when the constant term is not $0$.
