Find the sum to n terms of the series :`1xx2xx3 + 2xx3xx4 + 3xx4xx5 + dot dot dot`

Explanation

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

### Step 1: Identify the nth term of the series
The nth term of the series can be expressed as:
\[ T_n = n \times (n + 1) \times (n + 2) \]
This is because:
- For \(n = 1\), \(T_1 = 1 \times 2 \times 3\)
- For \(n = 2\), \(T_2 = 2 \times 3 \times 4\)
- For \(n = 3\), \(T_3 = 3 \times 4 \times 5\)

### Step 2: Expand the nth term
Now, we expand \(T_n\):
\[ T_n = n(n + 1)(n + 2) \]
Expanding this gives:
\[ T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n \]

### Step 3: Sum the series
We need to find the sum \(S_n\) of the first n terms:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k) \]
This can be split into three separate summations:
\[ S_n = \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k \]

### Step 4: Use known formulas for summations
We use the following formulas for the sums:
1. \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\)
2. \(\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}\)
3. \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n + 1)}{2}\right)^2\)

Substituting these 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 5: Simplify the expression
Now we simplify each term:
1. The first term becomes:
\[ \frac{n^2(n + 1)^2}{4} \]
2. The second term becomes:
\[ \frac{n(n + 1)(2n + 1)}{2} \]
3. The third term becomes:
\[ n(n + 1) \]

Combining these, we have:
\[ S_n = \frac{n^2(n + 1)^2}{4} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1) \]

### Step 6: Factor out common terms
Factoring out \(\frac{n(n + 1)}{4}\):
\[ S_n = \frac{n(n + 1)}{4} \left(n(n + 1) + 2(2n + 1) + 4\right) \]

### Step 7: Combine and simplify
Now we simplify the expression inside the brackets:
\[ n(n + 1) + 4n + 2 + 4 = n^2 + n + 4n + 6 = n^2 + 5n + 6 \]

Thus, we have:
\[ S_n = \frac{n(n + 1)(n^2 + 5n + 6)}{4} \]

### Final Result
The sum to n terms of the series is:
\[ S_n = \frac{n(n + 1)(n^2 + 5n + 6)}{4} \]