Find the sum to n terms of the series `1xx2+2xx3+ 3xx4+ 4xx5+......`

Explanation

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

### Step 1: Identify the nth term
The nth term of the series can be expressed as:
\[ T_n = n \times (n + 1) \]

### Step 2: Write the sum of the series
The sum of the first n terms, denoted as \(S_n\), can be written as:
\[ S_n = \sum_{k=1}^{n} k \times (k + 1) \]

### Step 3: Expand the summation
We can expand the term inside the summation:
\[ S_n = \sum_{k=1}^{n} (k^2 + k) \]
This can be separated into two summations:
\[ S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 4: Use summation formulas
We will use the known formulas for the summations:
1. The sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
2. The sum of the squares of the first n natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Substitute the formulas into the expression for \(S_n\)
Substituting these formulas into our expression for \(S_n\):
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 6: Simplify the expression
To combine the two terms, we need a common denominator. The common denominator is 6:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]
Combining the fractions:
\[ S_n = \frac{n(n + 1)(2n + 1 + 3)}{6} \]
This simplifies to:
\[ S_n = \frac{n(n + 1)(2n + 4)}{6} \]

### Step 7: Factor out common terms
Factoring out the common term:
\[ S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} \]
This can be simplified further:
\[ S_n = \frac{n(n + 1)(n + 2)}{3} \]

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

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