Question

The sum of all the products of the first $n$ natural numbers taken two at a time is

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

Answer 1

The correct option is A

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

Explanation for correct option:

Let the numbers are $x_1,x_2,....,x_n$

Therefore,

${\left(x_1+x_2+...+x_n\right)}^2=x_1^2+x_2^2+...+x_n^2+$$2$(Sum of product of numbers taken two at a time)

We know that Sum of first $n$ natural number is $\frac{n\left(n+1\right)}{2}$ and the sum of the square of first $n$ natural number is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$\Rightarrow$$2$(Sum of product of numbers taken two at a time)$={\left(\frac{n\left(n+1\right)}{2}\right)}^2-\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$\Rightarrow$Sum of product of numbers taken two at a time $=\frac{n^2{\left(n+1\right)}^2}{8}-\frac{n\left(n+1\right)\left(2n+1\right)}{12}$

$$\begin{gathered} =\frac{n\left(n+1\right)}{24}\left[3n\left(n+1\right)-2\left(2n+1\right)\right] \\ =\frac{n\left(n+1\right)}{24}\left[3n^2+3n-4n-2\right] \\ =\frac{n\left(n+1\right)}{24}\left[3n^2-n-2\right] \\ =\frac{n\left(n+1\right)}{24}\left[3n^2-3n+2n-2\right] \\ =\frac{n\left(n+1\right)}{24}\left[3n\left(n-1\right)+2\left(n-1\right)\right] \\ =\frac{n\left(n+1\right)}{24}\left[\left(3n+2\right)\left(n-1\right)\right] \end{gathered}$$

Hence, option (A) is correct.