The weighted means of of first n natural numbers whose weights are equal to the squares of corresponding numbers is

Options

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

Explanation

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

### Step 1: Define the Variables
Let \( x_i \) be the first \( n \) natural numbers:
\[ x_1 = 1, x_2 = 2, x_3 = 3, \ldots, x_n = n \]
The weights \( w_i \) are the squares of these numbers:
\[ w_1 = 1^2, w_2 = 2^2, w_3 = 3^2, \ldots, w_n = n^2 \]

### Step 2: Write the Formula for Weighted Mean
The formula for the weighted mean \( \bar{x} \) is given by:
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i w_i}{\sum_{i=1}^{n} w_i} \]

### Step 3: Calculate the Numerator
The numerator \( \sum_{i=1}^{n} x_i w_i \) can be expressed as:
\[ \sum_{i=1}^{n} x_i w_i = 1 \cdot 1^2 + 2 \cdot 2^2 + 3 \cdot 3^2 + \ldots + n \cdot n^2 \]
This can be simplified to:
\[ \sum_{i=1}^{n} i \cdot i^2 = \sum_{i=1}^{n} i^3 \]
The formula for the sum of cubes is:
\[ \sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

### Step 4: Calculate the Denominator
The denominator \( \sum_{i=1}^{n} w_i \) is:
\[ \sum_{i=1}^{n} w_i = 1^2 + 2^2 + 3^2 + \ldots + n^2 \]
The formula for the sum of squares is:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 5: Substitute the Values into the Weighted Mean Formula
Now substituting the values into the weighted mean formula:
\[ \bar{x} = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2} = \frac{\left( \frac{n(n+1)}{2} \right)^2}{\frac{n(n+1)(2n+1)}{6}} \]

### Step 6: Simplify the Expression
Simplifying the expression:
\[ \bar{x} = \frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2 \cdot 6}{4 \cdot n(n+1)(2n+1)} \]
Cancelling \( n(n+1) \) from numerator and denominator:
\[ \bar{x} = \frac{6n(n+1)}{4(2n+1)} = \frac{3n(n+1)}{2(2n+1)} \]

### Final Result
Thus, the weighted mean of the first \( n \) natural numbers with weights equal to the squares of the corresponding numbers is:
\[ \bar{x} = \frac{3n(n+1)}{2(2n+1)} \]