1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)=14n(n+1)(n+2)(n+3)


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

1.2.3 + 2.3.4 + 3.4.5 + ... + n(n+1)(n+2) = 14n(n+1)(n+2)(n+3)

Explore a challenging sequence sum problem where the series 1*2*3 + 2*3*4 + ... + n(n+1)(n+2) is set equal to 14n(n+1)(n+2)(n+3). This solution involves sequence patterns, polynomial summation, and algebraic techniques. Ideal for students in Grades 11-12.

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