Non - one numbers
If x is a non-real number such that x 5 = 1 , then evaluate ( n = 0 ∑ 2 0 1 5 x n ) 2 .
The answer is 1.
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2 solutions
x 5 = 1 = > x 5 − 1 = 0 = > ( x − 1 ) ( x 4 + x 3 + x 2 + x + 1 ) = 0 = > x 4 + x 3 + x 2 + x = − 1 because x isn't 1 . So x + x 2 + x 3 + x 4 + x 5 = 0 = > ( ∑ n = 0 2 0 1 5 x n ) 2 = ( x 0 + 0 ) 2 = 1
We have a geometric progression inside the parenthesis, hence the sum inside is given by: x − 1 x 2 0 1 6 − 1 = x − 1 x 2 0 1 5 ⋅ x − 1 = x − 1 ( x 5 ) 4 0 3 ⋅ x − 1 = x − 1 x − 1 = 1 because the only real solution of x 5 = 1 is 1 then, ( ∑ i = 0 2 0 1 5 x n ) 2 = 1 2 = 1