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Question

The sum of all possible products of the first $$n$$ natural numbers taken two by two is :


A
124n(n+1)(n1)(3n+2)
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B
n(n+1)(2n+1)6
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C
n(n+1)(2n1)(n+3)24
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D
None of these
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Solution

The correct option is A $$\dfrac {1}{24} n (n + 1) (n - 1)(3n + 2)$$
We have,
$$ ( a_1 + a_2 + ... a_n)^2 = \sum_{i=1}^n a_i^2 + 2 \sum_{ i \infty j} a_i a_j $$
Putting $$ a_1 = 1 , a_2 = 2 , ... a_n = n, $$ we get 
$$ 
\left[ \dfrac {n(n+1)}{2} \right]^2 = \dfrac {n(n+1)(2n+1)}{6} {6} + 2S $$
$$ \Rightarrow S = \dfrac {1}{24} n(n+1)(n-1)(3n+2) $$

Mathematics

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