No longer 2014
Let be a polynomial of degree .
Suppose
for all
Find
The answer is 0.
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1 solution
Let Q ( n ) = ( 1 + n ) P ( n ) − n . Then Q ( n ) has roots 0 , 1 , 2 , . . . , 2 0 1 0 .
Note that Q ( n ) has a degree of 2 0 1 1 . Hence Q ( n ) = k n ( n − 1 ) ( n − 2 ) ( n − 3 ) . . . ( n − 2 0 1 0 ) .
We only need to determine k . Note that Q ( − 1 ) = − 1 , so k ( − 1 ) ( − 2 ) ( − 3 ) . . . ( − 2 0 1 1 ) = 1 .
− ( 2 0 1 1 ! ) k = 1 ⟹ k = − 2 0 1 1 ! 1 .
Hence Q ( 2 0 1 2 ) = − 2 0 1 1 ! 1 ( 2 0 1 2 ) ( 2 0 1 1 ) . . . ( 2 ) = − 2 0 1 2 = 2 0 1 3 P ( 2 0 1 2 ) − 2 0 1 2 .
Then P ( 2 0 1 2 ) = 0 .