If the line `sqrt(5)x=y` meets the lines x = 1, x = 2, ….. x = n at points `A_(1), A_(2), ……, A_(n)` respectively, then `(0A_(1))^(2)+(0A_(2))^(2)+….+(0A_(n))^(2)` is equal to

Options

A. `3n^(2)+3n`
B. `2n^(3)+3n^(2)+n`
C. `3n^(3)+3n^(2)+2`
D. `(3//2)(n^(4)+2n^(3)+n^(3))`

Answer

To solve the problem, we need to find the sum of the squares of the distances from the origin to the points where the line \( \sqrt{5}x = y \) intersects the vertical lines \( x = 1, x = 2, \ldots, x = n \).

### Step-by-Step Solution:

1. **Identify the intersection points**:
The line \( \sqrt{5}x = y \) can be rewritten as \( y = \sqrt{5}x \). The intersection points with the lines \( x = k \) (where \( k = 1, 2, \ldots, n \)) can be found by substituting \( x = k \) into the equation of the line:
\[ y = \sqrt{5}k \]
Therefore, the points of intersection are:
\[ A_k = (k, \sqrt{5}k) \quad \text{for } k = 1, 2, \ldots, n \]

2. **Calculate the distance from the origin to each point**:
The distance \( OA_k \) from the origin \( O(0, 0) \) to the point \( A_k(k, \sqrt{5}k) \) is given by the distance formula:
\[ OA_k = \sqrt{(k - 0)^2 + (\sqrt{5}k - 0)^2} = \sqrt{k^2 + 5k^2} = \sqrt{6k^2} = k\sqrt{6} \]

3. **Square the distances**:
Now, we need to find \( OA_k^2 \):
\[ OA_k^2 = (k\sqrt{6})^2 = 6k^2 \]

4. **Sum the squares of the distances**:
We need to find the sum of the squares of the distances from the origin to all points \( A_1, A_2, \ldots, A_n \):
\[ \sum_{k=1}^{n} OA_k^2 = \sum_{k=1}^{n} 6k^2 = 6 \sum_{k=1}^{n} k^2 \]

5. **Use the formula for the sum of squares**:
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting this into our equation gives:
\[ \sum_{k=1}^{n} OA_k^2 = 6 \cdot \frac{n(n + 1)(2n + 1)}{6} = n(n + 1)(2n + 1) \]

### Final Answer:
Thus, the final result for \( (0A_1)^2 + (0A_2)^2 + \ldots + (0A_n)^2 \) is:
\[ n(n + 1)(2n + 1) \]