`(11^2 + 12^2 + 13^2 + …+ 20^2) = ?`

Options

A. 385
B. 2485
C. 2870
D. 3255

Answer

To solve the problem of finding the sum of squares from \(11^2\) to \(20^2\), we can use the formula for the sum of squares of the first \(n\) natural numbers, which is:

\[ S(n) = \frac{n(n + 1)(2n + 1)}{6} \]

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

We first need to calculate the sum of squares from \(1^2\) to \(20^2\). Here, \(n = 20\).

\[ S(20) = \frac{20(20 + 1)(2 \times 20 + 1)}{6} \]

Calculating each part:

- \(20 + 1 = 21\)
- \(2 \times 20 + 1 = 41\)

Now substituting these values into the formula:

\[ S(20) = \frac{20 \times 21 \times 41}{6} \]

Calculating the product:

\[ 20 \times 21 = 420 \]
\[ 420 \times 41 = 17220 \]

Now divide by 6:

\[ S(20) = \frac{17220}{6} = 2870 \]

### Step 2: Calculate the sum of squares from \(1^2\) to \(10^2\)

Next, we calculate the sum of squares from \(1^2\) to \(10^2\). Here, \(n = 10\).

\[ S(10) = \frac{10(10 + 1)(2 \times 10 + 1)}{6} \]

Calculating each part:

- \(10 + 1 = 11\)
- \(2 \times 10 + 1 = 21\)

Now substituting these values into the formula:

\[ S(10) = \frac{10 \times 11 \times 21}{6} \]

Calculating the product:

\[ 10 \times 11 = 110 \]
\[ 110 \times 21 = 2310 \]

Now divide by 6:

\[ S(10) = \frac{2310}{6} = 385 \]

### Step 3: Calculate the sum of squares from \(11^2\) to \(20^2\)

Now, to find the sum from \(11^2\) to \(20^2\), we subtract the sum from \(1^2\) to \(10^2\) from the sum from \(1^2\) to \(20^2\):

\[ S(11 \text{ to } 20) = S(20) - S(10) = 2870 - 385 \]

Calculating this gives:

\[ S(11 \text{ to } 20) = 2870 - 385 = 2485 \]

### Final Answer

Thus, the sum \(11^2 + 12^2 + 13^2 + \ldots + 20^2\) is:

\[ \boxed{2485} \]