
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 $x \neq 0$, we can find the sum of the first $n$ terms, $S_n = 1 + 2x + 3x^2 + \ldots nx^{n-1}$ ($n \in \mathbb{N}^*$) of the sequence $\{nx^{n-1}\}$ by using the "staggered subtraction method" of sequence summation. It can also be derived by summing the geometric series $x + x^2 + x^3 + \ldots + x^n$ ($n \in \mathbb{N}^*$) to get $x + x^2 + x^3 + \ldots + x^n = \frac{x - x^{n+1}}{1-x}$. Differentiating both sides as functions of $x$, $(x + x^2 + x^3 + \ldots + x^n)' = \left(\frac{x - x^{n+1}}{1-x}\right)'$, we obtain: $S_n = 1 + 2x + 3x^2 + \ldots nx^{n-1} = \frac{1-(n+1)x^n + nx^{n+1}}{(1-x)^2}$. Similarly, starting from the binomial expansion $(1+x)^n = 1 + {C_n^1}x + {C_n^2}x^2 + \ldots + {C_n^n}x^n$, derive the sum $S_n = 1 \times 2 \times {C_n^1} + 2 \times 3 \times {C_n^2} + 3 \times 4 \times {C_n^3} + \ldots + n \times (n+1) \times {C_n^n}$ ($n \geq 4$) and give the simplified result.

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