HMMT 二月 2003 · 代数 · 第 8 题

英文解析

8. Find the value of + + + · · · .
   2 2 2
   3 +1 4 +2 5 +3
   Solution: 13 / 36
   Each term takes the form
   1 1 = .
   n + ( n − 2) ( n + 2) · ( n − 1)2
   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 + ) = .
   3 4 36
   ( )2
   2 n 1