If the variate of a distribution takes the values `1^(2), 2^(2), 3^(2),….n^(2)` with frequencies `n, n-1, n-2,….3,2,1` respectively, then the mean value of the distribution is

Options

A. `(n(n+2))/(3)`
B. `(n(n+1)(n+2))/(6)`
C. `(n+2)/(3)`
D. `((n+1)(n+2))/(6)`

Answer

To find the mean value of the distribution given the variate values and their corresponding frequencies, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the variate values and frequencies**:
- The variate values are \(1^2, 2^2, 3^2, \ldots, n^2\).
- The corresponding frequencies are \(n, n-1, n-2, \ldots, 2, 1\).

2. **Calculate the mean formula**:
- The mean \( \mu \) of a distribution can be calculated using the formula:
\[ \mu = \frac{\sum f_i x_i}{\sum f_i} \]
where \( f_i \) is the frequency and \( x_i \) is the variate.

3. **Calculate \( \sum f_i x_i \)**:
- We need to compute \( \sum f_i x_i \):
\[ \sum f_i x_i = n \cdot 1^2 + (n-1) \cdot 2^2 + (n-2) \cdot 3^2 + \ldots + 1 \cdot n^2 \]
- This can be expressed as:
\[ \sum_{r=1}^{n} (n - r + 1) r^2 \]
- Rearranging gives:
\[ = \sum_{r=1}^{n} (n + 1 - r) r^2 = (n + 1) \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r^3 \]

4. **Use summation formulas**:
- The formulas for the summations are:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]
\[ \sum_{r=1}^{n} r^3 = \left(\frac{n(n + 1)}{2}\right)^2 \]

5. **Substitute the formulas into the mean calculation**:
- Substitute the summation results into the expression for \( \sum f_i x_i \):
\[ \sum f_i x_i = (n + 1) \cdot \frac{n(n + 1)(2n + 1)}{6} - \left(\frac{n(n + 1)}{2}\right)^2 \]

6. **Calculate \( \sum f_i \)**:
- The total frequency \( \sum f_i \) is:
\[ \sum f_i = n + (n - 1) + (n - 2) + \ldots + 2 + 1 = \frac{n(n + 1)}{2} \]

7. **Combine results to find the mean**:
- Now substituting back into the mean formula:
\[ \mu = \frac{(n + 1) \cdot \frac{n(n + 1)(2n + 1)}{6} - \left(\frac{n(n + 1)}{2}\right)^2}{\frac{n(n + 1)}{2}} \]
- Simplifying this expression leads to:
\[ \mu = \frac{(n + 1)(n + 2)}{6} \]

### Final Answer:
The mean value of the distribution is:
\[ \mu = \frac{(n + 1)(n + 2)}{6} \]