The sum of the series `1.2 + 2.3+ 3.4+……..` up to 20 tems is

Answer

To find the sum of the series \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots \) up to 20 terms, we can follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:
\[ a_n = n(n + 1) \]
This represents the product of \( n \) and \( n + 1 \).

### Step 2: Rewrite the General Term
We can rewrite the general term:
\[ a_n = n^2 + n \]

### Step 3: Sum the Series
We need to find the sum of the first 20 terms:
\[ S_{20} = \sum_{n=1}^{20} a_n = \sum_{n=1}^{20} (n^2 + n) \]
This can be separated into two sums:
\[ S_{20} = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n \]

### Step 4: Use the Formulas for the Sums
We will use the formulas for the sums of the first \( n \) natural numbers and the sum of the squares of the first \( n \) natural numbers:
- The sum of the first \( n \) natural numbers:
\[ \sum_{n=1}^{N} n = \frac{N(N + 1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{n=1}^{N} n^2 = \frac{N(N + 1)(2N + 1)}{6} \]

### Step 5: Calculate Each Sum
For \( N = 20 \):
1. Calculate \( \sum_{n=1}^{20} n^2 \):
\[ \sum_{n=1}^{20} n^2 = \frac{20(20 + 1)(2 \cdot 20 + 1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} \]
Calculating this gives:
\[ = \frac{20 \cdot 21 \cdot 41}{6} = 2870 \]

2. Calculate \( \sum_{n=1}^{20} n \):
\[ \sum_{n=1}^{20} n = \frac{20(20 + 1)}{2} = \frac{20 \cdot 21}{2} = 210 \]

### Step 6: Combine the Results
Now we can combine the results:
\[ S_{20} = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n = 2870 + 210 = 3080 \]

### Final Answer
Thus, the sum of the series up to 20 terms is:
\[ \boxed{3080} \]