Find the sum to n terms of the series (1.2.3) + (2.3.4) + (3.4.5) ...`

Answer

To find the sum to n terms of the series \( (1 \cdot 2 \cdot 3) + (2 \cdot 3 \cdot 4) + (3 \cdot 4 \cdot 5) + \ldots \), we can follow these steps:

### Step 1: Identify the general term of the series
The general term of the series can be expressed as:
\[ T_r = r \cdot (r + 1) \cdot (r + 2) \]
This can be expanded to:
\[ T_r = r(r^2 + 3r + 2) = r^3 + 3r^2 + 2r \]

### Step 2: Write the sum to n terms
The sum to n terms, \( S_n \), can be written as:
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (r^3 + 3r^2 + 2r) \]

### Step 3: Separate the summation
We can separate the summation into three parts:
\[ S_n = \sum_{r=1}^{n} r^3 + 3\sum_{r=1}^{n} r^2 + 2\sum_{r=1}^{n} r \]

### Step 4: Use summation formulas
We will use the following formulas for the summations:
- The sum of the first n natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
- The sum of the squares of the first n natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]
- The sum of the cubes of the first n natural numbers:
\[ \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

### Step 5: Substitute the formulas into the summation
Now we substitute these formulas into our expression for \( S_n \):
\[ S_n = \left( \frac{n(n+1)}{2} \right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} \]

### Step 6: Simplify the expression
Now we simplify each term:
1. The first term:
\[ \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{4} \]
2. The second term:
\[ 3 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{2} \]
3. The third term:
\[ 2 \cdot \frac{n(n+1)}{2} = n(n+1) \]

### Step 7: Combine all terms
Combining all terms gives:
\[ S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) \]

### Step 8: Factor out common terms
We can factor out \( \frac{n(n+1)}{2} \):
\[ S_n = \frac{n(n+1)}{2} \left( \frac{n(n+1)}{2} + (2n + 1) + 2 \right) \]

### Step 9: Simplify the expression inside the parentheses
Simplifying the expression inside the parentheses:
\[ \frac{n(n+1)}{2} + 2n + 1 + 2 = \frac{n(n+1) + 4n + 4}{2} = \frac{n^2 + 5n + 4}{2} \]

### Final Expression for \( S_n \)
Thus, the final expression for the sum to n terms is:
\[ S_n = \frac{n(n+1)}{2} \cdot \frac{n^2 + 5n + 4}{2} = \frac{n(n+1)(n^2 + 5n + 4)}{4} \]