Question

The sum of all possible product of $1st\, n$, 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
$\left(b_{1} +b_{2}+b_{3} + ....+b_{n}\right)^{2} =b^{2}_{1} +b_{2}^{2} +.....+b^{2}_{n} +2\sum_{i \ne j}b_{i}b_{j}$
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)
$\therefore \sum^{\text{Product of numbers}}_{\text{taken two at a time}}$
$= \frac{1}{2} \left(1 +2 +3 +...+n\right)^{2} - \frac{1}{2} \sum\limits^{n}_{n = 1} n^{2}$
$= \frac{n^{2} \left(n +1\right)^{2}}{2 \times4}- \frac{n\left(n +1\right) \left(2n +1\right)}{2 \times6}= \frac{n \left(n^{2} -1\right) \left(3n +2\right)}{24}$