Find the sum `11^2-1^2+12^2-2^2+13^2-3^2+……+20^2-10^2`

Explanation

To solve the problem, we need to find the sum of the series:

\[ S = 11^2 - 1^2 + 12^2 - 2^2 + 13^2 - 3^2 + \ldots + 20^2 - 10^2 \]

### Step-by-Step Solution:

1. **Rewrite the Series**:
We can separate the positive and negative terms in the series:
\[ S = (11^2 + 12^2 + 13^2 + \ldots + 20^2) - (1^2 + 2^2 + 3^2 + \ldots + 10^2) \]

2. **Sum of Squares from 11 to 20**:
The sum of squares from \(11^2\) to \(20^2\) can be expressed using the formula for the sum of squares of the first \(n\) natural numbers:
\[ \text{Sum of squares from } 1 \text{ to } n = \frac{n(n + 1)(2n + 1)}{6} \]
Therefore, the sum of squares from \(1\) to \(20\) is:
\[ S_{20} = \frac{20(20 + 1)(2 \cdot 20 + 1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} \]

3. **Sum of Squares from 1 to 10**:
Similarly, the sum of squares from \(1\) to \(10\) is:
\[ S_{10} = \frac{10(10 + 1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} \]

4. **Calculate the Sums**:
Now we can calculate both sums:
\[ S_{20} = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870 \]
\[ S_{10} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385 \]

5. **Substituting Back into the Original Expression**:
Now substituting back into our expression for \(S\):
\[ S = S_{20} - S_{10} = 2870 - 385 = 2485 \]

6. **Final Calculation**:
Therefore, the final answer is:
\[ S = 2485 \]

### Final Answer:
The sum \( S = 2485 \).