If the variance of the first 50 odd natural numbers is `V_(1)` and the variance of next 50 odd natural numbers is `V_(2)`, then `V_(1)+V_(2)` is equal to

Answer

To solve the problem, we need to find the variances \( V_1 \) and \( V_2 \) of the first 50 odd natural numbers and the next 50 odd natural numbers respectively, and then compute \( V_1 + V_2 \).

### Step 1: Identify the first 50 odd natural numbers
The first 50 odd natural numbers are:
\[ 1, 3, 5, 7, \ldots, 99 \]

### Step 2: Calculate the mean of the first 50 odd natural numbers
The mean \( \mu_1 \) can be calculated as:
\[ \mu_1 = \frac{\text{Sum of the first 50 odd natural numbers}}{50} \]
The sum of the first \( n \) odd natural numbers is given by \( n^2 \).
Thus, for \( n = 50 \):
\[ \text{Sum} = 50^2 = 2500 \]
So,
\[ \mu_1 = \frac{2500}{50} = 50 \]

### Step 3: Calculate the variance \( V_1 \)
The variance \( V_1 \) is calculated using the formula:
\[ V_1 = \frac{\sum (x_i^2)}{n} - \mu_1^2 \]
Where \( x_i \) are the first 50 odd natural numbers.

#### Step 3.1: Calculate \( \sum (x_i^2) \)
We need to calculate:
\[ \sum_{i=1}^{50} (2i - 1)^2 = \sum_{i=1}^{50} (4i^2 - 4i + 1) \]
This can be broken down into:
\[ 4\sum_{i=1}^{50} i^2 - 4\sum_{i=1}^{50} i + \sum_{i=1}^{50} 1 \]
Using the formulas:
- \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \)

For \( n = 50 \):
\[ \sum_{i=1}^{50} i^2 = \frac{50 \cdot 51 \cdot 101}{6} = 42925 \]
\[ \sum_{i=1}^{50} i = \frac{50 \cdot 51}{2} = 1275 \]
\[ \sum_{i=1}^{50} 1 = 50 \]
Putting it all together:
\[ \sum_{i=1}^{50} (2i - 1)^2 = 4 \cdot 42925 - 4 \cdot 1275 + 50 = 171700 - 5100 + 50 = 166650 \]

#### Step 3.2: Substitute into the variance formula
Now substituting back to find \( V_1 \):
\[ V_1 = \frac{166650}{50} - 50^2 = 3333 - 2500 = 833 \]

### Step 4: Identify the next 50 odd natural numbers
The next 50 odd natural numbers are:
\[ 101, 103, 105, \ldots, 199 \]

### Step 5: Calculate the mean of the next 50 odd natural numbers
The mean \( \mu_2 \) can be calculated as:
\[ \mu_2 = \frac{\text{Sum of the next 50 odd natural numbers}}{50} \]
The sum of these numbers is:
\[ \text{Sum} = 101 + 103 + \ldots + 199 = 50^2 + 100 \cdot 50 = 2500 + 5000 = 7500 \]
Thus,
\[ \mu_2 = \frac{7500}{50} = 150 \]

### Step 6: Calculate the variance \( V_2 \)
By the same reasoning as before, since the next 50 odd natural numbers are just the first 50 odd natural numbers shifted by 100, the variance \( V_2 \) will be the same as \( V_1 \):
\[ V_2 = V_1 = 833 \]

### Step 7: Calculate \( V_1 + V_2 \)
\[ V_1 + V_2 = 833 + 833 = 1666 \]

Thus, the final answer is:
\[ \boxed{1666} \]