Question

The mean (average) of the product of n natural numbers taken two at a time is

• A. $\frac{n(n+1)(3n+2)}{24}$
• B. $\frac{(n+1)(3n+1)}{24}$
• C. $\frac{(n+1)(3n+2)}{12}$
• D. None of these

Answer 1

The correct option is C $\frac{(n+1)(3n+2)}{12}$

Number of numbers from $1,2,...n$ taken two at a time are
${}^n{C}_2$ and for sum of the product of first n natural
numbers taken two at a time ,we consider ${(1+2+3+...n)}^2=1^2+2^2+3^2+...$ $+n^2+2\sum \sum a_i{a}_j$
$\therefore 2\sum \sum a_i{a}_j={(1+2+3+...n)}^2$$-(1^2+2^2+3^2+...+n^2)$
$=\frac{n^2{(n+1)}^2}{4}-\frac{n(n+1)(2n+1)}{6}$ $=\frac{(n-1)\left(n\right)(n+1)(3n+2)}{24}$
$\therefore$ Mean of the product of first $n$
Natural taken two at a time $=\frac{\text{Sum of the product}}{\text{total number of item}}$ $=\frac{(n-1)\left(n\right)(n+1)(3n+2)}{24{\times }^n{C}_2}=\frac{(n+1)(3n+2)}{12}$