If `sum` n = 210, then `sum n^(2) =`

Options

A. 2100
B. 2870
C. 2970
D. none

Answer

To solve the problem, we need to find the sum of the squares of the first n natural numbers given that the sum of the first n natural numbers equals 210.

### Step-by-Step Solution:

1. **Understand the given information**: We know that the sum of the first n natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
We are given that \( S_n = 210 \).

2. **Set up the equation**:
\[ \frac{n(n + 1)}{2} = 210 \]
To eliminate the fraction, multiply both sides by 2:
\[ n(n + 1) = 420 \]

3. **Rearrange the equation**:
\[ n^2 + n - 420 = 0 \]
This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \).

4. **Use the quadratic formula**:
The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -420 \).

5. **Calculate the discriminant**:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-420) = 1 + 1680 = 1681 \]

6. **Find the roots**:
\[ n = \frac{-1 \pm \sqrt{1681}}{2 \cdot 1} \]
Since \( \sqrt{1681} = 41 \):
\[ n = \frac{-1 \pm 41}{2} \]
This gives us two potential solutions:
\[ n = \frac{40}{2} = 20 \quad \text{and} \quad n = \frac{-42}{2} = -21 \]
Since n must be a positive integer, we take \( n = 20 \).

7. **Calculate the sum of squares**:
The formula for the sum of the squares of the first n natural numbers is:
\[ S_{n^2} = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting \( n = 20 \):
\[ S_{n^2} = \frac{20(20 + 1)(2 \cdot 20 + 1)}{6} \]
Simplifying this:
\[ = \frac{20 \cdot 21 \cdot 41}{6} \]

8. **Calculate the product**:
First, calculate \( 20 \cdot 21 = 420 \).
Then, calculate \( 420 \cdot 41 = 17220 \).
Finally, divide by 6:
\[ S_{n^2} = \frac{17220}{6} = 2870 \]

### Final Answer:
The sum of the squares of the first n natural numbers when \( n = 20 \) is:
\[ \boxed{2870} \]