If the roots of the polynomial:
P ( x ) = x 1 0 0 7 + x 1 0 0 6 + x 1 0 0 5 … ⋯ + x 2 + x + 1
are a 1 , a 2 , a 3 , a 4 … … a 1 0 0 7 , what is the value of i = 1 ∑ 1 0 0 7 F ( a i ) when
F ( x ) = x 2 0 1 5 + x 2 0 1 4 + x 2 0 1 3 + x 2 0 1 2 + … ⋯ + x 2 + x + 1 ?
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first note that P ( a i ) = 0 for i ∈ [ 1 , 2 , 3 … … , 1 0 0 7 ] and that F ( x ) = x 1 0 0 8 ( x 1 0 0 7 + x 1 0 0 6 + … … x + 1 ) + x 1 0 0 7 + x 1 0 0 6 + … … x + 1 F ( x ) = x 1 0 0 8 ( P ( x ) ) + P ( x ) F ( a i ) = x 1 0 0 8 ( P ( a i ) ) + P ( a i ) = x 2 0 1 3 ( 0 ) + 0 = 0 now add it 1012 times i = 1 ∑ 1 0 0 6 F ( a i ) = 0 × 1 0 0 7 = 0
Given that you changed the problem significantly, can you update your solution too? IE there is no a 1 0 1 2 .
I fail to understand what your F ( x ) in the solution is, and how it corresponds to the polynomial given in the question.
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sir, F ( x ) is defined at the end of the question . sorry for the change, there was a silly mistake in the degrees.
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1) Can you explain how your solution
F ( x ) = x 1 0 0 8 ( x 2 0 0 7 + x 2 0 0 6 + … … x + 1 ) + x 2 0 0 7 + x 2 0 0 6 + … … x + 1
corresponds with the problem
F ( x ) = x 2 0 1 5 + x 2 0 1 4 + x 2 0 1 3 + x 2 0 1 2 + … ⋯ + x 2 + x + 1
2) Can you explain how your solution
F ( x ) = x 1 0 0 8 ( x 2 0 0 7 + x 2 0 0 6 + … … x + 1 ) + x 2 0 0 7 + x 2 0 0 6 + … … x + 1
is equal to F ( x ) = x 1 0 0 8 P ( x ) + P ( x ) ?
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@Calvin Lin – sorry again, it shoud be 1007 in the solution, not 2007
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Note that P ( x ) = x − 1 x 1 0 0 8 − 1 , so a i 1 0 0 8 = 1 for all i .
Then F ( a i ) = a i − 1 a i 2 0 1 6 − 1 = a i − 1 ( a i 1 0 0 8 − 1 ) ( a i 1 0 0 8 + 1 ) = 0 for all i , so i = 1 ∑ 1 0 0 7 F ( a i ) = 0 .