Find the sum to `n` terms of the series `1 *2 *3 + 2*4*6 + 3*6*9+...`

Answer

To find the sum of the series \(1 \times 2 \times 3 + 2 \times 4 \times 6 + 3 \times 6 \times 9 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the General Term
The series consists of terms of the form:
- The first term is \(1 \times 2 \times 3\)
- The second term is \(2 \times 4 \times 6\)
- The third term is \(3 \times 6 \times 9\)

We can express the general term \(T_n\) for the \(n\)-th term as:
\[ T_n = n \times (2n) \times (3n) = 6n^3 \]

### Step 2: Write the Sum of the First \(n\) Terms
The sum of the first \(n\) terms \(S_n\) can be written as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} 6k^3 \]

### Step 3: Use the Formula for the Sum of Cubes
We know that the sum of the first \(n\) cubes is given by the formula:
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
Thus, we can substitute this into our expression for \(S_n\):
\[ S_n = 6 \sum_{k=1}^{n} k^3 = 6 \left(\frac{n(n+1)}{2}\right)^2 \]

### Step 4: Simplify the Expression
Now, simplifying the expression:
\[ S_n = 6 \cdot \frac{n^2(n+1)^2}{4} = \frac{3}{2} n^2 (n+1)^2 \]

### Final Result
Thus, the sum of the first \(n\) terms of the series is:
\[ S_n = \frac{3}{2} n^2 (n+1)^2 \]