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

Explanation

To find the sum \( 11^2 - 1^2 + 12^2 - 2^2 + 13^2 - 3^2 + \ldots + 20^2 - 10^2 \), we can break it down into two separate sums:

1. The sum of squares from \( 11^2 \) to \( 20^2 \)
2. The sum of squares from \( 1^2 \) to \( 10^2 \)

### Step 1: Calculate the sum of squares from \( 11^2 \) to \( 20^2 \)

The sum of squares from \( n^2 \) to \( m^2 \) can be calculated using the formula:

\[ \text{Sum} = \sum_{k=n}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} - \frac{(n-1)n(2(n-1)+1)}{6} \]

For our case, \( n = 11 \) and \( m = 20 \).

Calculating \( \sum_{k=1}^{20} k^2 \):

\[ \sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} \]

Calculating \( \sum_{k=1}^{10} k^2 \):

\[ \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} \]

### Step 2: Substitute values into the formula

Calculating \( \sum_{k=1}^{20} k^2 \):

\[ \sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870 \]

Calculating \( \sum_{k=1}^{10} k^2 \):

\[ \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385 \]

### Step 3: Find the difference

Now, we can find the required sum:

\[ \sum_{k=11}^{20} k^2 = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{10} k^2 \]

Substituting the values we calculated:

\[ \sum_{k=11}^{20} k^2 = 2870 - 385 = 2485 \]

### Step 4: Final calculation

Now, we can find the final sum:

\[ \text{Final Sum} = \sum_{k=11}^{20} k^2 - \sum_{k=1}^{10} k^2 = 2485 - 385 = 2100 \]

Thus, the final answer is:

\[ \boxed{2100} \]