HMMT 二月 2003 · 代数 · 第 8 题

HMMT February 2003 — Algebra — Problem 8

解析

Find the value of + + + · · · . 2 2 2 3 +1 4 +2 5 +3 Solution: 13 / 36 Each term takes the form 1 1 = . 2 n + ( n − 2) ( n + 2) · ( n − 1) Using the method of partial fractions, we can write (for some constants A, B ) 1 A B = + ( n + 2) · ( n − 1) ( n + 2) ( n − 1) ⇒ 1 = A · ( n − 1) + B · ( n + 2) 1 1 Setting n = 1 we get B = , and similarly with n = − 2 we get A = − . Hence the 3 3 sum becomes [( ) ( ) ( ) ( ) ] 1 1 1 1 1 1 1 1 1 · − + − + − + − + · · · . 3 2 5 3 6 4 7 5 8 1 1 Thus, it telescopes, and the only terms that do not cancel produce a sum of · ( + 3 2 1 1 13