To expand $\frac{1}{(1 - x)^{3n + 3}}$, we can apply the binomial series expansion for expressions of the form $(1 - x)^{-k}$, where $k$ is a positive integer or a real number.

The binomial series expansion for $(1 - x)^{-k}$ is given by:

$(1 - x)^{-k} = \sum_{m=0}^{\infty} \binom{k+m-1}{m} x^m$

In this case, we have $k = 3n + 3$, so the expansion becomes:

$\frac{1}{(1 - x)^{3n+3}} = \sum_{m=0}^{\infty} \binom{3n+3+m-1}{m} x^m$

Step-by-step expansion:

1. The general term of the expansion is:

   $\text{Term} = \binom{3n+2+m}{m} x^m$

   Here, $\binom{3n+2+m}{m}$ is the binomial coefficient.

2. The first few terms of the expansion are:

   $\frac{1}{(1 - x)^{3n+3}} = 1 + (3n+3) x + \frac{(3n+3)(3n+4)}{2!} x^2 + \frac{(3n+3)(3n+4)(3n+5)}{3!} x^3 + \cdots$

Conclusion:

This gives the infinite series representation of $\frac{1}{(1 - x)^{3n+3}}$. Each term involves increasingly larger powers of $x$ and binomial coefficients based on $3n + 3$.

Would you like more details or clarifications?

5 Related Questions:

1. How is the binomial expansion applied to negative exponents?
2. Can you show the first four terms of the expansion for a specific value of $n$?
3. How does the convergence of this series depend on $x$?
4. What is the general term in a binomial expansion?
5. How can this series expansion be applied in practical problems?

Tip:

For series expansions, always check the convergence radius to ensure the expansion is valid for the values of $x$ you're working with.