Find the sum of squares of first 11 natural numbers.

Answer

To find the sum of squares of the first 11 natural numbers, we can use the formula for the sum of squares of the first n natural numbers:

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

where \( n \) is the number of natural numbers.

### Step-by-Step Solution:

1. **Identify the value of n**:
Here, we need to find the sum of squares for the first 11 natural numbers, so \( n = 11 \).

2. **Substitute n into the formula**:
Substitute \( n = 11 \) into the formula:
\[ S = \frac{11(11 + 1)(2 \cdot 11 + 1)}{6} \]

3. **Calculate \( n + 1 \) and \( 2n + 1 \)**:
- \( n + 1 = 11 + 1 = 12 \)
- \( 2n + 1 = 2 \cdot 11 + 1 = 22 + 1 = 23 \)

4. **Substitute these values back into the formula**:
Now, substitute \( n + 1 \) and \( 2n + 1 \) into the formula:
\[ S = \frac{11 \cdot 12 \cdot 23}{6} \]

5. **Calculate the product in the numerator**:
First, calculate \( 11 \cdot 12 \):
\[ 11 \cdot 12 = 132 \]
Now multiply this by 23:
\[ 132 \cdot 23 = 3036 \]

6. **Divide by 6**:
Now, divide the product by 6:
\[ S = \frac{3036}{6} = 506 \]

7. **Conclusion**:
Therefore, the sum of squares of the first 11 natural numbers is:
\[ S = 506 \]