The sum of all possible product of $1^{s t}$ $n$ natural numbers taken two at a time is

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

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

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\frac{n (n^{2} + 1) (3 n + 2)}{6}

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\frac{n (n^{2} - 1) (3 n + 2)}{6}

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Solution

The correct option is A $\frac{n (n^{2} - 1) (3 n + 2)}{24}$
Consider
${(b_{1} + b_{2} + b_{3} + . . . + b_{n})}_{1}^{2} = b_{1}^{2} + b_{2}^{2} + . . . + b_{n}^{2} + 2 \sum i = j b_{i} b_{j}$
taking $b_{1} = 1, b_{2} = 2, . . ., b_{n} = n$
$∴ {(1 + 2 + 3 + . . . + n)}^{2} = 1^{2} + 2^{2} + 3^{2} + . . . + n^{2} + 2 \sum$ (Product of number taken two at a time)
$\Rightarrow 2 \sum b_{i} b_{j} = {(1 + 2 + 3 + . . . + n)}^{2} - n \sum n = 1 n^{2} = \frac{n^{2} {(n + 1)}^{2}}{4} - \frac{n (n + 1) (2 n + 1)}{6} = \frac{n (n^{2} - 1) (3 n + 2)}{12}$
$\Rightarrow \sum b_{i} b_{j} = \frac{n (n^{2} - 1) (3 n + 2)}{24}$

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