Question

The sum of all possible products of the first $n$ natural numbers taken two by two is :

• A. $\frac{1}{24}n(n+1)(n-1)(3n+2)$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. $\frac{n(n+1)(2n-1)(n+3)}{24}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{24}n(n+1)(n-1)(3n+2)$

We have,
$(a_1+a_2+...a_n{)}^2={\sum }_{i=1}^n{a}_i^2+2{\sum }_{i\infty j}{a}_i{a}_j$
Putting $a_1=1,a_2=2,...a_n=n,$ we get
$
\left[ \dfrac {n(n+1)}{2} \right]^2 = \dfrac {n(n+1)(2n+1)}{6} {6} + 2S $
$\Rightarrow S=\frac{1}{24}n(n+1)(n-1)(3n+2)$