The weighted mean of first n natural numbers whose weights are equal to the corresponding number is equal to

Options

A. `1/3(2n+1)`
B. `1/2(n+4)`
C. `1/5(2n+3)`
D. none of these

Answer

To find the weighted mean of the first \( n \) natural numbers where the weights are equal to the corresponding numbers, we can follow these steps:

### Step 1: Define the Variables
Let the first \( n \) natural numbers be \( x_1, x_2, \ldots, x_n \) where \( x_i = i \) for \( i = 1, 2, \ldots, n \). The weights \( w_i \) are also equal to the corresponding numbers, so \( w_i = i \).

### Step 2: Calculate the Weighted Mean Formula
The formula for the weighted mean is given by:

\[ \text{Weighted Mean} = \frac{w_1x_1 + w_2x_2 + \ldots + w_nx_n}{w_1 + w_2 + \ldots + w_n} \]

### Step 3: Substitute the Values
Substituting the values of \( w_i \) and \( x_i \):

\[ \text{Weighted Mean} = \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + \ldots + n \cdot n}{1 + 2 + 3 + \ldots + n} \]

### Step 4: Simplify the Numerator
The numerator can be expressed as:

\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 \]

The formula for the sum of the squares of the first \( n \) natural numbers is:

\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Simplify the Denominator
The denominator is the sum of the first \( n \) natural numbers:

\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 6: Substitute Back into the Weighted Mean Formula
Now substituting these results back into the weighted mean formula:

\[ \text{Weighted Mean} = \frac{\frac{n(n + 1)(2n + 1)}{6}}{\frac{n(n + 1)}{2}} \]

### Step 7: Simplify the Expression
Now we simplify the expression:

\[ \text{Weighted Mean} = \frac{n(n + 1)(2n + 1)}{6} \times \frac{2}{n(n + 1)} = \frac{2n + 1}{3} \]

### Final Result
Thus, the weighted mean of the first \( n \) natural numbers whose weights are equal to the corresponding numbers is:

\[ \text{Weighted Mean} = \frac{2n + 1}{3} \]

---