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PUMaC 2010 · 代数（A 组） · 第 6 题

PUMaC 2010 — Algebra (Division A) — Problem 6

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Assume that f ( a + b ) = f ( a ) + f ( b ) + ab , and that f (75) − f (51) = 1230. Find f (100). √ ◦ ◦ ◦ ◦ 50
解析
  1. Assume that f ( a + b ) = f ( a ) + f ( b ) + ab , and that f (75) − f (51) = 1230. Find f (100). Solution: One has that f (0) = f (0) + f (0), so f (0) = 0. Moreover, f ( n + 1) = f ( n ) + f (1) + n , n − 1 ∑ n ( n − 1) so that f ( n ) = ( i + f (1)) = + nf (1). Plugging in n = 75 and n = 51, one gets 2 i =0 45 2775 − 1275 + 24 f (1) = 1230, so f (1) = − . Thus, f (100) = 3825. 4 √ ◦ ◦ ◦ ◦ 50