作者Honor1984 (希望願望成真)
看板trans_math
標題Re: [極限]
時間Mon Sep 29 02:48:36 2008
※ 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
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◆ From: 122.124.101.106
→ Qmmm:謝謝~ 202.151.60.129 09/29 11:19
推 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