Question

Sum of all the products of the first n positive integers taken two at a time is?

• A. $\frac{1}{24}(n-1)n(n+1)(3n+2)$
• B. $\frac{1}{48}(n-2)n(n-1)n^2$
• C. $\frac{1}{6}(n+1)(n+2)(n+5)$
• D. None of these

Answer 1

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

${(x_1+x_2+x_3+......x_n)}^2$

$=(x_1^2+x_2^2+x_3^2+......x_n^2)+2$(Sum of product of numbers taken two at a time)

$\Rightarrow 2\times$ (Sum of product of numbers taken two at a time)

$={\left[\frac{n(n+1)}{2}\right]}^2-\frac{n(n+1)(2n+1)}{6}$

$\Rightarrow$ (Sum of product of numbers taken two at a time)$\frac{1}{4}n(n+1)[\frac{n(n+1)}{2}-\frac{2n+1}{3}]$

$=\frac{1}{24}n(n+1)(n-1)(3n+2)$

Hence the correct answer is $\frac{1}{24}n(n+1)(n-1)(3n+2)$