A simple cubic sum
Let function f ( x ) = 3 1 x 3 − 2 1 x 2 + 3 x − 1 2 5 . What is the value of i = 1 ∑ 2 0 1 7 f ( 2 0 1 8 i ) ?
The answer is 2017.
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2 solutions
k = 1 ∑ 2 0 1 7 f ( 2 0 1 8 k ) = k = 1 ∑ 2 0 1 7 ( 3 1 ( 2 0 1 8 k ) 3 − 2 1 ( 2 0 1 8 k ) 2 + 3 ( 2 0 1 8 k ) − 1 2 5 ) = 3 ⋅ 2 0 1 8 3 1 k = 1 ∑ 2 0 1 7 k 3 − 2 ⋅ 2 0 1 8 2 1 k = 1 ∑ 2 0 1 7 k 2 + 2 0 1 8 3 k = 1 ∑ 2 0 1 7 k − 1 2 5 k = 1 ∑ 2 0 1 7 1 = 4 ⋅ 3 ⋅ 2 0 1 8 3 2 0 1 7 2 ⋅ 2 0 1 8 2 − 6 ⋅ 2 ⋅ 2 0 1 8 2 2 0 1 7 ⋅ 2 0 1 8 ⋅ 4 0 3 5 + 2 ⋅ 2 0 1 8 3 ⋅ 2 0 1 7 ⋅ 2 0 1 8 − 1 2 5 ⋅ 2 0 1 7 = 1 2 ⋅ 2 0 1 8 2 0 1 7 2 − 1 2 ⋅ 2 0 1 8 2 0 1 7 ⋅ 4 0 3 5 + 2 3 ⋅ 2 0 1 7 − 1 2 5 ⋅ 2 0 1 7 = 1 2 ⋅ 2 0 1 8 2 0 1 7 ( 2 0 1 7 − 4 0 3 5 ) + 2 3 ⋅ 2 0 1 7 − 1 2 5 ⋅ 2 0 1 7 = − 1 2 2 0 1 7 + 1 2 1 8 ⋅ 2 0 1 7 − 1 2 5 ⋅ 2 0 1 7 = 1 2 1 2 ⋅ 2 0 1 7 = 2 0 1 7
f ( x ) + f ( 1 − x ) = 2
Ans:- 2 ( 1 0 0 8 ) + f ( 2 0 1 8 1 0 0 9 )
From the first equation, f ( 2 1 ) = 1
Thus, the answer is 2 0 1 6 + 1 = 2 0 1 7