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# The weighted mean of the first n natural numbers when their weights are equal to the corresponding natural number is : A.$\dfrac{{{\text{n}}\,{\text{ + }}\,{\text{1}}}}{{\text{2}}}$ B.$\dfrac{{{\text{2n}}\,{\text{ + }}\,{\text{1}}}}{{\text{3}}}$ C.$\dfrac{{{\text{(n}}\,{\text{ + }}\,{\text{1)}}\,{\text{(2n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{6}}}$ D. None of these  Hint: A weighted mean is a kind of average. Instead of each data point contributing equally to the final mean, some data points contribute more “weight” than others.

Weighted means are a simple concept . All you have to do is multiply the numbers in your data set by the corresponding weights , and sum these multiplied numbers up , then , you need to add up all the weights and divide the previous sum by this sum .
Let ‘s make a table .

 Data set weight 12::n 12::n

According to the question , the weight of a natural number is equal to the number itself .
Now , first we do $\sum\limits_{}^{} {{\text{Data}}\,{\text{set}}\,{{ \times }}\,{\text{weight}}}$
Which is ${{\text{1}}^{\text{2}}}\,{\text{ + }}\,{{\text{2}}^{\text{2}}}\,{\text{ + }}\,{{\text{3}}^{\text{2}}}\,{\text{ + }}..........\,{\text{ + }}\,{{\text{n}}^{\text{2}}}$ .
The standard summation of square of n natural number is given by :
$\sum\limits_{{\text{m = 1}}}^{\text{n}} {\,{{\text{m}}^{\text{2}}}\,{\text{ = }}\,\dfrac{{{\text{n(n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{6}}}} {\text{(2n}}\,{\text{ + }}\,{\text{1)}}$
Then we sum up the weights which is
${\text{1}}\,{\text{ + }}\,{\text{2}}\,{\text{ + }}\,{\text{3}}\,{\text{ + }}\,.........\,{\text{ + }}\,{\text{n}}$
The sum of the first n natural numbers is given by $\sum\limits_{{\text{m = 1}}}^{\text{n}} {\,{\text{m}}\,{\text{ = }}\,\dfrac{{{\text{n(n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{2}}}}$
Now the weighted mean
$\begin{gathered} {\text{M}}\,{\text{ = }}\,\dfrac{{\sum\limits_{{\text{m}}\,{\text{ = }}\,{\text{1}}}^{\text{n}} {{{\text{m}}^{\text{2}}}} }}{{\sum\limits_{{\text{m}}\,{\text{ = }}\,{\text{1}}}^{\text{n}} {\text{m}} }} \\ {\text{ = }}\,\dfrac{{\dfrac{{{\text{n(n}}\,{\text{ + }}\,{\text{1)(2n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{6}}}}}{{\dfrac{{{\text{n(n + 1)}}}}{{\text{2}}}}} \\ {\text{ = }}\,\dfrac{{{\text{2n + 1}}}}{{\text{3}}} \\ \end{gathered}$

Hence, the option B is correct .

Note: Remember the following formulae :
1)Sum of first n natural no. s = $\dfrac{{{\text{n(n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{2}}}$
2)Sum of squares of first n natural no. s = $\dfrac{{{\text{n}}\,{\text{(n}}\,{\text{ + }}\,{\text{1)}}\,{\text{(2n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{6}}}\,$
3)Sum of cubes of first n natural no. s = ${\left( {\dfrac{{{\text{n}}\,{\text{(n}}\,{\text{ + }}\,{\text{1)}}}}{{\text{2}}}} \right)^{\text{2}}}$
These formulae will come in handy in the future .
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