Question

The mean (average) of the product of n natural numbers taken two at a time is:
(a). ${\displaystyle \frac{n(n+1)(3n+2)}{24}}$
(b). ${\displaystyle \frac{(n+1)(3n+1)}{24}}$
(c). ${\displaystyle \frac{(n+1)(3n+1)}{12}}$
(d). None of these

Answer 1

Hint: The formula for mean is the division of the sum of all the observations to the total number of observations. Now, the number of possibilities for the product of n natural numbers taken two at a time is selecting 2 natural numbers from n natural numbers. And the sum of the observations (i.e. product of n natural numbers taken two at a time) is calculated as follows: by multiplying the sum of first n natural numbers by itself and then equating them with the sum of the square of each n natural number with the twice of the sum of the product of n natural numbers taken two at a time. Solving this equation will give you the sum of the product of n natural numbers taken two at a time.

Complete step-by-step solution:
We have to find the mean (average) of the product of n natural numbers taken two at a time.
We know the formula for mean of any observations as:
$Mean={\displaystyle \frac{\text{Sum of observations}}{\text{Total number of observations}}}$
Total number of observations includes the number of ways of writing the product of n natural numbers taken two at a time which will be selecting 2 natural numbers from n natural numbers.
${}^n{C}_2$
Sum of observation includes the sum of product of n natural numbers taken two at a time are:
$(1+2+3+.....+n)(1+2+3+....+n)=1^2+2^2+3^2+.....+n^2+2\sum _{i,j=1}^n{a}_i{a}_j$…….. Eq. (1)
We know the sum of first n natural numbers and sum of square of first n natural numbers which we have shown below.
$\begin{array}{rl}& 1+2+3+....+n={\displaystyle \frac{n(n+1)}{2}}\\ & 1^2+2^2+3^2+.....+n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}\end{array}$
Substituting the above values in eq. (1) we get,
$\begin{array}{rl}& {\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}+2\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2-{\displaystyle \frac{n(n+1)(2n+1)}{6}}=2\sum _{i,j=1}^n{a}_i{a}_j\end{array}$
Taking $$n(n+1)$$ as common in the above equation we get,
$\begin{array}{rl}& n(n+1)({\displaystyle \frac{n(n+1)}{4}}-{\displaystyle \frac{2n+1}{6}})=2\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\displaystyle \frac{n(n+1)}{2}}\left({\displaystyle \frac{3n(n+1)-4n-2}{12}}\right)=\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\displaystyle \frac{n(n+1)}{2}}\left({\displaystyle \frac{3n^2+3n-4n-2}{6}}\right)=\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\displaystyle \frac{n(n+1)}{2}}\left({\displaystyle \frac{3n^2-n-2}{6}}\right)=\sum _{i,j=1}^n{a}_i{a}_j\end{array}$
We can factorize $3n^2-n-2$ as follows:
$\begin{array}{rl}& 3n^2-n-2\\ & =3n^2-3n+2n-2\\ & =3n(n-1)+2(n-1)\\ & =(3n+2)(n-1)\end{array}$
Substituting the above factorization we get,
$$\begin{array}{rl}& {\displaystyle \frac{n(n+1)}{2}}\left({\displaystyle \frac{3n^2-n-2}{6}}\right)=\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\displaystyle \frac{n(n+1)}{2}}\left({\displaystyle \frac{(3n+2)(n-1)}{6}}\right)=\sum _{i,j=1}^n{a}_i{a}_j\\ & \Rightarrow {\displaystyle \frac{n(n+1)}{12}}\left((3n+2)(n-1)\right)=\sum _{i,j=1}^n{a}_i{a}_j\end{array}$$
Now, substituting sum of observations as $${\displaystyle \frac{n(n+1)}{12}}\left((3n+2)(n-1)\right)$$ and number of observations as ${}^n{C}_2$ in the mean formula we get,
$Mean={\displaystyle \frac{{\displaystyle \frac{n(n+1)}{12}}\left((3n+2)(n-1)\right)}{{}^n{C}_2}}$
We can write ${}^n{C}_2={\displaystyle \frac{n(n-1)}{2}}$ in the above formula.
$Mean={\displaystyle \frac{{\displaystyle \frac{n(n+1)}{12}}\left((3n+2)(n-1)\right)}{{\displaystyle \frac{n(n-1)}{2}}}}$
In the above formula, ${\displaystyle \frac{n(n-1)}{2}}$ will be cancelled out and we are left with:
$$Mean={\displaystyle \frac{(n+1)(3n+2)}{6}}$$
Hence, the correct option is (c).

Note: The possible mistake that could happen in the above problem is that you might think that the sum of the product of n natural numbers taken two at a time is the sum of the square of first n natural numbers.
The sum of the square of first n natural numbers is:
$1^2+2^2+3^2+.....+n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}$
The above interpretation of the sum of the product of n natural numbers taken two at a time is wrong because it is given that we are taking any two natural numbers at a time not specifically two same natural numbers at a time.