HMMT 二月 2005 · 代数 · 第 8 题

英文解析

8. Compute
   ∞
   ∑
   .n
   4 2
   n + n + 1
   n =0
   Solution: 1 / 2
   Note that
   4 2 4 2 2 2 2 2 2 2
   n + n + 1 = ( n + 2 n + 1) − n = ( n + 1) − n = ( n + n + 1)( n − n + 1) .
   Decomposing into partial fractions, we find that
   ( )
   n 1 1 1 = − .
   4 2 2 2
   n + n + 1 2 n − n + 1 n + n + 1
   1 1 1
   Now, note that if f ( n ) = , then f ( n + 1) = = . It follows
   2 2 2
   n − n +1 ( n +1) − ( n +1)+1 n + n +1
   that
   ∞
   ( )
   ∑
   n 1 = ( f (0) − f (1)) + ( f (1) − f (2)) + ( f (2) − f (3)) + · · · .
   4 2
   n + n + 1 2
   n =0
   Since f ( n ) tends towards 0 as n gets large, this sum telescopes to f (0) / 2 = 1 / 2.