HMMT 十一月 2020 · 冲刺赛 · 第 30 题

HMMT November 2020 — Guts Round — Problem 30

[15] Let a , a , a , . . . be a sequence of positive real numbers that satisfies 1 2 3 ( ) ∞ ∑ n 1 a = , n k k 5 n = k a for all positive integers k . The value of a − a + a − a + · · · can be expressed as , where a, b are 1 2 3 4 b relatively prime positive integers. Compute 100 a + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#

解析

[15] Let a , a , a , . . . be a sequence of positive real numbers that satisfies 1 2 3 ( ) ∞ ∑ n 1 , a = n k k 5 n = k a for all positive integers k . The value of a − a + a − a + · · · can be expressed as , where a, b are 1 2 3 4 b relatively prime positive integers. Compute 100 a + b . Proposed by: Akash Das Answer: 542 1 Solution: Let S = . In order to get the coefficient of a to be − 1, we need to have S − 3 S . k k 2 1 3 5 This subtraction makes the coefficient of a become − 6. Therefore, we need to add 7 S to make the 3 3 coefficient of a equal to 1. The coefficient of a in S − 3 S + 7 S is 14, so we must subtract 15 S . 4 4 1 3 5 4 We can continue to pattern to get that we want to compute S − 3 S + 7 S − 15 S + 31 S − · · · . To 1 2 3 4 5 prove that this alternating sum equals a − a + a − a + · · · , it suffices to show 1 2 3 4 ( ) n ∑ n i i i +1 ( − ( − 2) + ( − 1) ) = ( − 1) . i i =1 i i i +1 To see this is true, note that the left hand side equals − (1 − 2) + (1 − 1) = ( − 1) by binomial expansion. (We may rearrange the sums since the positivity of the a ’s guarantee absolute convergence.) i Now, all that is left to do is to compute ∞ ∞ ∞ − 1 − 2 i i − 1 i i ∑ ∑ ∑ (2 − 1)( − 1) ( − 1) ( − 2) 5 5 5 = − = − = . − 1 − 2 i i i 5 5 5 42 1 − 1 − 5 5 i =1 i =1 i =1