**f(x)=x$^3$ln(1+2x)**

Write the first four non-zero terms of the Taylor Series for the above function with x centered at a=0.

Using this model:

**ln(1+x) = Σ$\frac{(-1)^{k}(x)^{k}}{k}$**

I get the following...

x$^{3}$ln(1+2x) => x$^3$Σ$\frac{(-1)^{k}(2x)^{k}}{k}$

k=0 -----> Cannot divide by zero.

k=1 -----> x$^3$$\frac{(-1)^{1}(2x)^{1}}{1}$$\frac{(x-a)}{1!}$ -----> (0)$^3$$\frac{(-1)^{1}(2(0))^{1}}{1}$$\frac{(x-0)}{1!}$ = 0

k=2 -----> x$^3$$\frac{(-1)^{2}(2x)^{2}}{2}$$\frac{(x-a)}{1!}$ -----> (0)$^3$$\frac{(-1)^{2}(2(0))^{2}}{1}$$\frac{(x-0)}{2!}$ = 0

and so on...

I keep getting zero for every value of k in the series so I never reach a non-zero term.

I don't even know if I am doing this wrong but I am assuming that I am supposed to get an answer that isn't, "there are no non-zero terms" for this expansion.