
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
When $x \neq 1$ and $x \neq 0$, the sum of the first $n$ terms of the sequence $\{nx^{n-1}\}$, denoted as $S_n = 1 + 2x + 3x^2 + \ldots + nx^{n-1} (n \in \mathbb{N}^*)$, can be obtained using the "displacement subtraction method" for sequence summation. It can also be derived from the sum of the geometric sequence $x + x^2 + x^3 + \ldots + x^n (n \in \mathbb{N}^*)$, which is first calculated as $x + x^2 + x^3 + \ldots + x^n = \frac{x - x^{n+1}}{1-x}$. Both sides are functions of $x$, and by differentiating both sides simultaneously, $(x + x^2 + x^3 + \ldots + x^n)' = \left( \frac{x - x^{n+1}}{1-x} \right)'$, thus obtaining $S_n = 1 + 2x + 3x^2 + \ldots + nx^{n-1} = \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2}$. Using the same method, starting from the binomial expansion $(1+x)^n = 1 + C_n^1x + C_n^2x^2 + \ldots + C_n^nx^n$, it can be found that $S_n = 1 \times 2 \times C_n^1 + 2 \times 3 \times C_n^2 + 3 \times 4 \times C_n^3 + \ldots + n(n+1) \times C_n^n (n \geqslant 4)$ equals to \_\_\_\_\_\_\_\_. (Please fill in the simplest result).

Present the answer in LaTex format: \boxed{Your answer}
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