To find the value of $n$ for which the sum of the sequence

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + n(n+1)(n+2) = 14n(n+1)(n+2)(n+3)$

we will proceed by investigating the pattern of the sequence and examining if there is a general formula that matches the form given on the right-hand side.

Step 1: Understanding the Pattern of Terms in the Sum

The $k$-th term of the sequence can be written as: $T_k = k \cdot (k+1) \cdot (k+2)$ where $k$ starts from $1$ up to $n$.

The sum of the sequence up to $n$ terms is: $S_n = \sum_{k=1}^n k \cdot (k+1) \cdot (k+2)$

Step 2: Analyzing the Right Side

We are given that this sum equals: $S_n = 14n(n+1)(n+2)(n+3)$

Step 3: Checking Values of $n$

To solve this equation, we can substitute small values of $n$ to check if both sides are equal.

1. For $n = 1$: $S_1 = 1 \cdot 2 \cdot 3 = 6$ Right side: $14 \cdot 1 \cdot 2 \cdot 3 \cdot 4 = 336$ This does not hold, so $n \neq 1$.

2. For $n = 2$: $S_2 = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 = 6 + 24 = 30$ Right side: $14 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 1680$ This does not hold either, so $n \neq 2$.

3. For $n = 3$: [ S_3 = 1 \cdot 2 \cdot 3 + 2 \cdot