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

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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