The simple and weighted arithmetic mean of the first n natural numbers, the weights being the corresponding numbers is

Options

A. `(n+1)/(2), (2n+1)/(2)`
B. `(n+1)/(2), (2n+1)/(4)`
C. `(n+1)/(2), (2n+1)/(3)`
D. `n, n^(2)`

Answer

To solve the problem of finding the simple and weighted arithmetic mean of the first \( n \) natural numbers, where the weights are the corresponding numbers, we will follow these steps:

### Step 1: Calculate the Simple Arithmetic Mean
The simple arithmetic mean (AM) of the first \( n \) natural numbers is given by the formula:

\[ \text{AM} = \frac{\sum_{i=1}^{n} i}{n} \]

The sum of the first \( n \) natural numbers can be calculated using the formula:

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

Substituting this into the AM formula gives:

\[ \text{AM} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

### Step 2: Calculate the Weighted Arithmetic Mean
The weighted arithmetic mean (WAM) is calculated using the formula:

\[ \text{WAM} = \frac{\sum_{i=1}^{n} i \cdot w_i}{\sum_{i=1}^{n} w_i} \]

In this case, the weights \( w_i \) are the corresponding numbers themselves, so \( w_i = i \). Therefore, we need to calculate:

\[ \text{WAM} = \frac{\sum_{i=1}^{n} i \cdot i}{\sum_{i=1}^{n} i} \]

This simplifies to:

\[ \text{WAM} = \frac{\sum_{i=1}^{n} i^2}{\sum_{i=1}^{n} i} \]

The sum of the squares of the first \( n \) natural numbers is given by:

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

Thus, substituting this into the WAM formula gives:

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

### Final Result
So, the simple arithmetic mean of the first \( n \) natural numbers is:

\[ \text{AM} = \frac{n + 1}{2} \]

And the weighted arithmetic mean of the first \( n \) natural numbers, with weights being the corresponding numbers, is:

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