If `A_(n)=(n, n+1)`, then `10(A_(10)A_(11))^(2)+11(A_(11)A_(12))^(2)+….+20(A_(20)A_(21))^(2)` is equal to

Answer

To solve the problem, we need to evaluate the expression:

\[ 10(A_{10}A_{11})^2 + 11(A_{11}A_{12})^2 + \ldots + 20(A_{20}A_{21})^2 \]

where \( A_n = (n, n+1) \).

### Step-by-Step Solution:

1. **Identify the Points**:
- We have \( A_n = (n, n+1) \) and \( A_{n+1} = (n+1, n+2) \).
- Therefore, the coordinates of the points are:
- \( A_{10} = (10, 11) \)
- \( A_{11} = (11, 12) \)
- \( A_{12} = (12, 13) \)
- ...
- \( A_{20} = (20, 21) \)

2. **Calculate the Distance Squared**:
- The distance \( A_n A_{n+1} \) is calculated as:
\[ A_n A_{n+1} = \sqrt{(n+1 - n)^2 + (n+2 - (n+1))^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]
- Therefore, \( (A_n A_{n+1})^2 = 2 \).

3. **Substituting into the Expression**:
- Now, substituting this into the original expression:
\[ 10(A_{10}A_{11})^2 + 11(A_{11}A_{12})^2 + \ldots + 20(A_{20}A_{21})^2 = 10 \cdot 2 + 11 \cdot 2 + 12 \cdot 2 + \ldots + 20 \cdot 2 \]
- This simplifies to:
\[ 2(10 + 11 + 12 + \ldots + 20) \]

4. **Calculating the Sum**:
- The sum \( 10 + 11 + 12 + \ldots + 20 \) is an arithmetic series.
- The number of terms \( n \) in this series is \( 20 - 10 + 1 = 11 \).
- The sum of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (a + l) \]
where \( a \) is the first term, \( l \) is the last term, and \( n \) is the number of terms.
- Here, \( a = 10 \), \( l = 20 \), and \( n = 11 \):
\[ S = \frac{11}{2} \times (10 + 20) = \frac{11}{2} \times 30 = 11 \times 15 = 165 \]

5. **Final Calculation**:
- Now substituting back into our expression:
\[ 2 \times 165 = 330 \]

Thus, the final result is:

\[ \boxed{330} \]