Question

The mean (average) of the product of n natural numbers taken two at a time is:
(a). $\dfrac{n\left( n+1 \right)\left( 3n+2 \right)}{24}$
(b). $\dfrac{\left( n+1 \right)\left( 3n+1 \right)}{24}$
(c). $\dfrac{\left( n+1 \right)\left( 3n+1 \right)}{12}$
(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:
$Mean=\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$
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:
$\left( 1+2+3+.....+n \right)\left( 1+2+3+....+n \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}+2\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}}$…….. 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.
$\begin{align}
  & 1+2+3+....+n=\dfrac{n\left( n+1 \right)}{2} \\
 & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\
\end{align}$
Substituting the above values in eq. (1) we get,
$\begin{align}
  & {{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+2\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow {{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}-\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}=2\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
\end{align}$
Taking \[n\left( n+1 \right)\] as common in the above equation we get,
$\begin{align}
  & n\left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{4}-\dfrac{2n+1}{6} \right)=2\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{2}\left( \dfrac{3n\left( n+1 \right)-4n-2}{12} \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{2}\left( \dfrac{3{{n}^{2}}+3n-4n-2}{6} \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{2}\left( \dfrac{3{{n}^{2}}-n-2}{6} \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
\end{align}$
We can factorize $3{{n}^{2}}-n-2$ as follows:
$\begin{align}
  & 3{{n}^{2}}-n-2 \\
 & =3{{n}^{2}}-3n+2n-2 \\
 & =3n\left( n-1 \right)+2\left( n-1 \right) \\
 & =\left( 3n+2 \right)\left( n-1 \right) \\
\end{align}$
Substituting the above factorization we get,
\[\begin{align}
  & \dfrac{n\left( n+1 \right)}{2}\left( \dfrac{3{{n}^{2}}-n-2}{6} \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{2}\left( \dfrac{\left( 3n+2 \right)\left( n-1 \right)}{6} \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
 & \Rightarrow \dfrac{n\left( n+1 \right)}{12}\left( \left( 3n+2 \right)\left( n-1 \right) \right)=\sum\limits_{i,j=1}^{n}{{{a}_{i}}{{a}_{j}}} \\
\end{align}\]
Now, substituting sum of observations as \[\dfrac{n\left( n+1 \right)}{12}\left( \left( 3n+2 \right)\left( n-1 \right) \right)\] and number of observations as ${}^{n}{{C}_{2}}$ in the mean formula we get,
$Mean=\dfrac{\dfrac{n\left( n+1 \right)}{12}\left( \left( 3n+2 \right)\left( n-1 \right) \right)}{{}^{n}{{C}_{2}}}$
We can write ${}^{n}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}$ in the above formula.
$Mean=\dfrac{\dfrac{n\left( n+1 \right)}{12}\left( \left( 3n+2 \right)\left( n-1 \right) \right)}{\dfrac{n\left( n-1 \right)}{2}}$
In the above formula, $\dfrac{n\left( n-1 \right)}{2}$ will be cancelled out and we are left with:
\[Mean=\dfrac{\left( n+1 \right)\left( 3n+2 \right)}{6}\]
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:
${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
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.