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※ 引述《Qmmm (Q蛆蛆)》之銘言: : 設a_1+ a_2+ ....+ a_n= 3n^2 : 求 lim [ √(a_2+ a_4+ ...+ a_(2n))- √(a_1+ a_3+ a_5+ ...+ a_(2n-1))]= ? : n→∞ 令A = a_2+ a_4+ ...+ a_(2n) B = a_1+ a_3+ a_5+ ...+ a_(2n-1) a_1 + ... + a_2n = 12n^2 = A + B a_n = 3[(n)^2 - (n-1)^2] = 3(2n-1) A = Σ3[2*2k-1] = 6(n+1)n - 3n = 6n^2 + 3n 求lim √A - √B = lim √A - √(12n^2-A) 2A -12n^2 = lim ------------ √A + √(12n^2-A) 6n = lim ------------ √A + √(12n^2-A) 6 = ----------- √6 + √6 = √6 /2 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.124.101.106
Qmmm:謝謝~ 202.151.60.129 09/29 11:19
andy2007:請問 a_1 + ... + a_2n = 12n^2 = A + B140.125.205.230 09/29 11:58
andy2007:和 a_n = 3[(n)^2 - (n-1)^2] = 3(2n-1)140.125.205.230 09/29 11:58
andy2007:這兩行是怎麼來的呢 ? 程度差 煩請指導140.125.205.230 09/29 11:59