Question

The sum of all possible product of $1stn$, natural numbers taken two at $a$ time is

• A. $\frac{n\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{24}$
• B. $\frac{n\left(n^2-1\right)\left(3n+2\right)}{24}$
• C. $\frac{n\left(n^2+1\right)\left(3n+2\right)}{6}$
• D. $\frac{n\left(n^2-1\right)\left(3n+2\right)}{6}$

Answer 1

Answer: (B)

Consider
${\left(b_1+b_2+b_3+....+b_n\right)}^2=b_1^2+b_2^2+.....+b_n^2+2{\sum }_{i\ne j}{b}_i{b}_j$
Taking $b_1,=1,b_2=2,...,b_n=n$
$\therefore {\left(1+2+3+...+n\right)}^2=1^2+2^2+3^2+...+n^2+2\Sigma$
(Product of numbers taken two at a time)
$\therefore {\sum }_{\text{taken two at a time}}^{\text{Product of numbers}}$
$=\frac{1}{2}{\left(1+2+3+...+n\right)}^2-\frac{1}{2}\sum _{n=1}^n{n}^2$
$=\frac{n^2{\left(n+1\right)}^2}{2\times 4}-\frac{n\left(n+1\right)\left(2n+1\right)}{2\times 6}=\frac{n\left(n^2-1\right)\left(3n+2\right)}{24}$