HMMT 二月 2005 · 代数 · 第 8 题
HMMT February 2005 — Algebra — Problem 8
题目详情
- Compute ∞ ∑ n . 4 2 n + n + 1 n =0
解析
- 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.