HMMT 二月 2006 · 代数 · 第 9 题

HMMT February 2006 — Algebra — Problem 9

Compute the value of the infinite series ∞ 4 2 ∑ n + 3 n + 10 n + 10 n 4 2 · ( n + 4) n =2

解析

Compute the value of the infinite series ∞ 4 2 ∑ n + 3 n + 10 n + 10 n 4 2 · ( n + 4) n =2 3 11 Answer: 10 Solution: We employ the difference of squares identity, uncovering the factorization 4 2 2 2 2 2 of the denominator: n + 4 = ( n + 2) − (2 n ) = ( n − 2 n + 2)( n + 2 n + 2). Now, 4 2 2 n + 3 n + 10 n + 10 3 n + 10 n + 6 = 1 + 4 4 n + 4 n + 4 4 1 = 1 + − 2 2 n − 2 n + 2 n + 2 n + 2 ∞ ∞ 4 2 ∑ ∑ n + 3 n + 10 n + 10 1 4 1 = ⇒ = + − n 4 n n 2 n 2 2 · ( n + 4) 2 2 · ( n − 2 n + 2) 2 · ( n + 2 n + 2) n =2 n =2 ∞ ∑ 1 1 1 = + − n − 2 2 n 2 2 2 · (( n − 1) + 1) 2 · (( n + 1) + 1) n =2 1 1 1 1 1 11 The last series telescopes to + , which leads to an answer of + + = . 2 10 2 2 10 10