The weighted arithmetic mean of the first n natural numbers whose weights are equal to the corresponding numbers is given by
(A) \[\dfrac{1}{2}\left( {n + 1} \right)\]
(B) \[\dfrac{1}{2}\left( {n + 2} \right)\]
(C) \[\dfrac{1}{3}\left( {2n + 1} \right)\]
(D) \[\dfrac{1}{3}n\left( {2n + 1} \right)\]

Hint: In this question, we have to find out the required value from the given particulars.
We need to first consider the formula of weighted arithmetic mean for first n natural numbers then putting equal weights as given in the question.
Solving the derived mean we will get the required solution.

Formula used:
i). The weighted mean of a non-empty finite multi-set of data \[\left\{ {{x_1},{x_2},{x_3},.....,{x_n}} \right\}\], with corresponding non-negative weights \[\left\{ {{w_1},{w_2},{w_3},.....{w_n}} \right\}\] is
\[\overline x = \dfrac{{{w_1}{x_1} + {w_2}{x_2} + {w_3}{x_3} + ..... + {w_n}{x_n}}}{{{w_1} + {w_2} + {w_3} + ....... + {w_n}}}\]
ii). The sum of first n natural numbers \[1,2,3,.....,n\] is \[\dfrac{{n\left( {n + 1} \right)}}{2}\]
iii). The sum of square of first n natural numbers \[1,2,3,.....,n\] is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]

Complete step by step solution:
We need to find out the weighted arithmetic mean of the first n natural numbers whose weights are equal to the corresponding numbers.
First n natural numbers are \[1,2,3,.....,n\] and their corresponding weights are \[1,2,3,.....,n\].
Now, using the formula for weighted mean of the first n natural numbers \[\left\{ {1,2,3,.....,n} \right\}\] with corresponding weights \[\left\{ {1,2,3,.....,n} \right\}\] is
\[\overline x = \dfrac{{1 \times 1 + 2 \times 2 + 3 \times 3 + ....... + n \times n}}{{1 + 2 + 3 + .... + n}}\]
\[ \Rightarrow \dfrac{{{1^2} + {2^2} + {3^2} + .... + {n^2}}}{{1 + 2 + 3 + .... + n}}\]
Now, we know that the sum of first n natural numbers \[1,2,3,.....,n\] is \[\dfrac{{n\left( {n + 1} \right)}}{2}\] and the sum of squares of first n natural numbers \[1,2,3,.....,n\] is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\].
Using this we get,
$\Rightarrow$ \[\overline x = \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{\dfrac{{n\left( {n + 1} \right)}}{2}}}\]
Simplifying we get,
\[ = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \times \dfrac{2}{{n\left( {n + 1} \right)}}\]
Hence,
\[ = \dfrac{{\left( {2n + 1} \right)}}{3}\]
Hence the required weighted arithmetic mean is \[\dfrac{{\left( {2n + 1} \right)}}{3}\].

$\therefore $ Option (C) is the correct option.

Note: The weighted arithmetic mean is similar to an ordinary arithmetic mean (the most common type of average), except that instead of each of the data points contributing equally to the final average, some data points contribute more than others. The notion of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics.
If all the weights are equal, then the weighted mean is the same as the arithmetic mean. While weighted means generally behave in a similar fashion to arithmetic means, they do have a few counterintuitive properties, as captured for instance in Simpson’s paradox.