Let f : N → R be a function such that
k = 0 ∑ n f ( k ) = 2 0 1 6 ( n + 2 ) ( n + 1 )
Find ( f ( 2 0 1 4 ) ) − 1 .
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k = 0 ∑ 2 0 1 4 f ( k ) = f ( 0 ) + f ( 1 ) + … + f ( 2 0 1 4 ) = 2 0 1 6 ( 2 0 1 4 + 2 ) ( 2 0 1 4 + 1 ) = 2 0 1 5 k = 0 ∑ 2 0 1 3 f ( k ) = f ( 0 ) + f ( 1 ) + … + f ( 2 0 1 3 ) = 2 0 1 6 ( 2 0 1 3 + 2 ) ( 2 0 1 3 + 1 ) = 2 0 1 5 2 0 1 6 ⋅ 2 0 1 4 f ( 2 0 1 4 ) = k = 0 ∑ 2 0 1 4 f ( k ) − k = 0 ∑ 2 0 1 3 f ( k ) = 2 0 1 5 − 2 0 1 5 2 0 1 6 ⋅ 2 0 1 4 = 2 0 1 5 − 2 0 1 5 ( 2 0 1 5 + 1 ) ( 2 0 1 5 − 1 ) = 2 0 1 5 2 0 1 5 2 − ( 2 0 1 5 2 − 1 2 ) = 2 0 1 5 1
( f ( 2 0 1 4 ) ) − 1 = ( 2 0 1 5 1 ) − 1 = 2 0 1 5
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Nice problem!
f ( n ) = k = 0 ∑ n f ( k ) − k = 0 ∑ n − 1 f ( k ) = 2 0 1 6 n + 2 n + 1 − 2 0 1 6 n + 1 n = ( n + 1 ) ( n + 2 ) 2 0 1 6 so f ( 2 0 1 4 ) = 2 0 1 5 1 and the reciprocal we seek is 2 0 1 5