How do you use the binomial series to expand #x^4/(1-3x)^3#?

This reduces to

To expand $\frac{x^4}{(1-3x)^3}$ using the binomial series, you can start by recognizing that $\frac{1}{1-3x}$ can be expressed as a geometric series using the formula $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$ when $|x| < 1$. We need to make $1-3x$ appear in this form.

First, rewrite the expression $\frac{1}{1-3x}$ as $(1-3x)^{-1}$. Now, using the binomial series expansion for $(1+x)^n$, where $n$ is a real number, we have $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$.

Applying this formula to $(1-3x)^{-1}$, we get $(1-3x)^{-1} = 1 + 3x + 3^2x^2 + 3^3x^3 + \ldots$ when $|3x| < 1$, which is equivalent to $|x| < \frac{1}{3}$.

Now, we can substitute this series expansion into $\frac{x^4}{(1-3x)^3}$:

Expanding this expression, we'll get terms of the form $x^k$ for various values of $k$. Since we're interested in the terms up to $x^4$, we'll only need to consider terms up to $x^3$ in each factor of the expansion.

After expanding and simplifying, the terms containing $x^4$ will be the coefficients of $x^4$ in the expansions of $(1+3x)^3$, $(1+3x)^2$, and $(1+3x)$, multiplied together with $x^4$.