The sum of the squares of the first n positive integers `1^2 + 2^2 + 3^2 + ………+ n^2` is `(n(n +1)(2n+1))/(6)`. What is the sum of the squares of the first 9 positive integers?

Options

A. 90
B. 125
C. 200
D. 285

Answer

To find the sum of the squares of the first 9 positive integers using the formula provided, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the formula**: The formula for the sum of the squares of the first \( n \) positive integers is given by:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]

2. **Set the value of \( n \)**: Since we want to find the sum of the squares of the first 9 positive integers, we set \( n = 9 \).

3. **Substitute \( n \) into the formula**:
\[ S_9 = \frac{9(9 + 1)(2 \cdot 9 + 1)}{6} \]

4. **Calculate \( n + 1 \) and \( 2n + 1 \)**:
- \( n + 1 = 9 + 1 = 10 \)
- \( 2n + 1 = 2 \cdot 9 + 1 = 18 + 1 = 19 \)

5. **Substitute these values back into the formula**:
\[ S_9 = \frac{9 \cdot 10 \cdot 19}{6} \]

6. **Calculate the product in the numerator**:
\[ 9 \cdot 10 = 90 \]
\[ 90 \cdot 19 = 1710 \]

7. **Divide by 6**:
\[ S_9 = \frac{1710}{6} = 285 \]

### Final Answer:
The sum of the squares of the first 9 positive integers is \( 285 \).

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