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Question: The mean (average) of the product of n natural numbers taken two at a time is: (a). \(\dfrac{n\lef...

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

Explanation

Solution

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=Sum of observationsTotal number of observationsMean=\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.
nC2{}^{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)=12+22+32+.....+n2+2∑i,j=1naiaj\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.
1+2+3+....+n=n(n+1)2 12+22+32+.....+n2=n(n+1)(2n+1)6 \begin{aligned} & 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{aligned}
Substituting the above values in eq. (1) we get,
(n(n+1)2)2=n(n+1)(2n+1)6+2∑i,j=1naiaj ⇒(n(n+1)2)2−n(n+1)(2n+1)6=2∑i,j=1naiaj \begin{aligned} & {{\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{aligned}
Taking n(n+1)n\left( n+1 \right) as common in the above equation we get,
n(n+1)(n(n+1)4−2n+16)=2∑i,j=1naiaj ⇒n(n+1)2(3n(n+1)−4n−212)=∑i,j=1naiaj ⇒n(n+1)2(3n2+3n−4n−26)=∑i,j=1naiaj ⇒n(n+1)2(3n2−n−26)=∑i,j=1naiaj \begin{aligned} & 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{aligned}
We can factorize 3n2−n−23{{n}^{2}}-n-2 as follows:
3n2−n−2 =3n2−3n+2n−2 =3n(n−1)+2(n−1) =(3n+2)(n−1) \begin{aligned} & 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{aligned}
Substituting the above factorization we get,

& \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{aligned}$$ Now, substituting sum of observations as $$\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.