A problem for a friend II
The roots of x 2 0 1 6 − x 2 0 1 5 + x 2 0 1 3 − x 2 0 1 0 + x 2 0 0 6 − … − 1 = 0 are r 1 , r 2 , … , r 2 0 1 6 .
Find
1 ≤ i < j ≤ 2 0 1 6 ∑ r i r j
The answer is 0.
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1 solution
Got tricked....
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I too got it right on the third try!
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I just thought of a question followup...
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@Julian Poon – Looking forward to it
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@Sualeh Asif – Just shared it here
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@Julian Poon – Sorry guys, the previous problem had an error. Heres the updated version:
I realised there was a problem when there were 4 attempts and no solvers, which is weird.
@Julian Poon – Share it! I'm looking forward to it~
Looking at the monic polynomial we see we want all permutations of the roots(both real and complex) taken two at a time.
This is equivalent to thd coefficient of the x n − 2 in an monic n -degree polynomial .This is the direct consequence of Vieta's formula .
∴ We can see that the coefficient of x 2 0 1 6 − 2 = x 2 0 1 4 is just 0 .
P.s. I believe @Jake Lai omitted the term on purpose to trick the solver.