
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 found using the "method of differences" for series summation. It can also be derived from the sum of the geometric series $x+x^{2}+x^{3}+\ldots+x^{n} (n \in \mathbb{N}^{*})$, which gives $x+x^{2}+x^{3}+\ldots+x^{n}= \frac{x-x^{n+1}}{1-x}$. Differentiating both sides with respect to $x$, we get $(x+x^{2}+x^{3}+\ldots+x^{n})'=( \frac{x-x^{n+1}}{1-x})'$, and thus $S_{n}=1+2x+3x^{2}+\ldots+nx^{n-1}= \frac{1-(n+1)x^{n}+nx^{n+1}}{(1-x)^{2}}$. Using the same method, starting from the binomial expansion $(1+x)^{n}=1+ C_{n}^{1}x+C_{n}^{2}x^{2}+\ldots+C_{n}^{n}x^{n}$, we can find that the sum $S_{n}=1\times2\times C_{n}^{1}+2\times3\times C_{n}^{2}+3\times4\times C_{n}^{3}+\ldots+n\times(n+1)\times C_{n}^{n} (n\geqslant 4)$ is \_\_\_\_\_\_ (Please fill in the simplest result).

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