Find the nth term and the sum to n terms of the series 1.2+ 2.3 +3.4 + ...

Answer

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

### Step 1: Identify the nth term of the series
The series can be expressed as:
\[ S_n = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n \cdot (n + 1) \]
The general term \(T_r\) (the rth term) of the series can be written as:
\[ T_r = r \cdot (r + 1) \]
This is because the first term corresponds to \(r = 1\), the second term to \(r = 2\), and so on.

### Step 2: Simplify the nth term
We can simplify the expression for \(T_r\):
\[ T_r = r^2 + r \]

### Step 3: Find the sum of the first n terms
To find the sum of the first n terms \(S_n\), we need to sum the terms from \(T_1\) to \(T_n\):
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (r^2 + r) \]
This can be separated into two sums:
\[ S_n = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

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

### Step 5: Substitute the formulas into the sum
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: Combine the terms
To combine these fractions, we need a common denominator, which is 6:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \]
This simplifies to:
\[ S_n = \frac{n(n + 1)(2n + 1 + 3)}{6} = \frac{n(n + 1)(2n + 4)}{6} \]

### Step 7: Factor out common terms
Factoring out the common terms gives us:
\[ S_n = \frac{n(n + 1)(n + 2)}{3} \]

### Final Results
Thus, the nth term \(T_n\) and the sum of the first n terms \(S_n\) are:
- **nth term**: \(T_n = n(n + 1)\)
- **Sum of n terms**: \(S_n = \frac{n(n + 1)(n + 2)}{3}\)