PUMaC 2022 · 团队赛 · 第 3 题
PUMaC 2022 — Team Round — Problem 3
题目详情
- Provided that { α } are the 28 distinct roots of 29 x + 28 x + . . . + 2 x + 1 = 0, then the i i =1 28 P p 1 absolute value of can be written as for relatively prime positive integers p, q . 2 (1 − α ) q i i =1 Find p + q .
解析
- Provided that { α } are the 28 distinct roots of 29 x + 28 x + . . . + 2 x + 1 = 0, then the i i =1 28 P p 1 absolute value of can be written as for relatively prime positive integers p, q . 2 (1 − α ) q i i =1 Find p + q . Proposed by Ben Zenker Answer: 275 1 Let n = 30, and let p ( x ) denote the given polynomial. Then are the roots of the function 1 − α i x − 1 1 x − 1 n − 2 p ( ). Therefore are the roots of the polynomial q ( x ) = x p ( ), which can be x 1 − α x i written as n − 2 X k n − 2 − k q ( x ) = ( k + 1)( x − 1) x k =0 n − 2 n − 3 n − 4 Let the three leading terms of q ( x ) be denoted ax + bx + cx . By Vieta’s formulas, 2 the desired sum is given by ( − b/a ) − 2( c/a ). n n − 2 − m m We claim that the coefficient of x is given as ( − 1) ( m + 1) . To see this, note that m +2 k k +1 n − 2 − m k n − 2 − k m m the coefficient of x in ( k + 1)( x − 1) x is ( k + 1)( − 1) = ( − 1) ( m + 1) m m +1 n m by the Binomial Theorem. Summing over m ≤ k ≤ n − 2, we find ( − 1) ( m + 1) by the m +2 Hockey-Stick Identity, as claimed. − 2 n ( n − 1)( n − 2) / 6 n n n 2 It follows that a = , b = − 2 , and c = 3 . Thus b/a = = − ( n − 2) 2 3 4 n ( n − 1) / 2 3 3 n ( n − 1)( n − 2)( n − 3) / 24 1 4 2 and c/a = = ( n − 2)( n − 3). The desired sum is therefore ( n − 2) − n ( n − 1) / 2 4 9 1 1 1 ( n − 2)( n − 3), which reduces to ( n − 2)[(8 n − 16) − (9 n − 27)] = ( n − 2)(11 − n ). Plugging 2 18 18 1 266 in n = 30, the sum of squares becomes (28)( − 19) = − . Thus p = 266 , q = 9 and our 18 9 answer is 266 + 9 = 275.