PUMaC 2020 · 代数（B 组） · 第 7 题
PUMaC 2020 — Algebra (Division B) — Problem 7
题目详情
- Suppose we have a sequence a , a , . . . of positive real numbers so that for each positive integer 1 2 ∑ n 2 √ n, we have that a a = n . Determine the first value of k so a > 100 . k k k =1 b k c 2 qiπ
解析
- Suppose we have a sequence a , a , . . . of positive real numbers so that for each positive integer 1 2 ∑ n 2 √ n, we have that a a = n . Determine the first value of k so a > 100 . k k k =1 b k c Proposed by: Frank Lu Answer: 1018 Note: On the original algebra test, we had forgotten to include the phrase “positive real num- bers.” 2 n − 1 Notice that this relation becomes the equation that a = , by subtracting this for n and n √ a b n c n − 1 . From here, to figure out when this is larger than 100 , we need to make some deductions about the rough behavior of this sequence. Notice here that, trying smaller values, we have that a = 3 , a = 5 , a = 7 / 3 , a = 3 , a = 11 / 3 . 2 3 4 5 6 √ 2 n − 1 √ √ √ √ √ First, notice that a = a > na . Observe then that for n = 1295 = n 2 b n c− 1 b b n cc b b n cc 4 6 − 1 , notice that a > 35 a = 105 , so hence our maximal value is going to be at most 1295 . 1295 5 √ √ In particular, we see that a for our maximal value is either going to be a , a , a , a , 1 2 3 4 b b n cc √ √ 7 2 n − 1 √ or a . But notice however that b b n cc = 5; if it is 4 , notice that this is at most < 5 3 2 b n c− 1 7 1250 8750 = < 100 . And furthermore, if it is less than 4 , we see that we can bound this more 3 31 93 √ √ √ 2 2 √ 2( b n c +1) − 1 2 b n c +4 b n c +1 7 / 2 2 n − 1 √ √ √ √ crudely by 5 < 5 = 5 = 5( b n c + 5 / 2 + ) . On the 2 b n c− 1 2 b n c− 1 2 b n c− 1 2 b n c− 1 √ one hand, we see that if b n c ≤ 3 , this is at most 5(3 + 5 / 2 + 7 / 2) < 45 . On the other hand, √ b n c ≥ 4 , so this is at most 5(15 + 5 / 2 + 1 / 2) < 90 < 100 . 6 n − 3 √ In particular, we require then that for our minimal value for n, we have that a = . n 2 b n c− 1 On one hand, again we can use our bounds above to see that this is bounded above by √ √ 7 / 2 7 / 2 √ √ 3( b n c + 5 / 2 + ); we therefore need to have that b n c + 5 / 2 + > 33 . But 2 b n c− 1 2 b n c− 1 √ √ with n ≥ 25 in this particular subcase, this means that we have that b n c + 3 > 33 , or that √ 6 n − 3 b n c > 30 . We start with this being 31; we then get that a = . To be larger than 100 , n 61 this requires that 6 n > 6103 , or that n ≥ 1018 . 2 qiπ