推 andy2007:原來如此 還有一個問題140.125.205.230 09/29 14:09
→ andy2007:A = Σ3[2*2k-1] = 6(n+1)n - 3n140.125.205.230 09/29 14:09
→ andy2007:那個是算偶數項的總和嗎?那個2K是...140.125.205.230 09/29 14:10
→ andy2007:最後n是怎麼來的? ...麻煩了Orz...140.125.205.230 09/29 14:11
→ Honor1984:是算偶數項總和沒錯 但我用2k表示,k為連122.124.105.166 09/29 14:34
→ Honor1984:續整數 只是計算方法的問題122.124.105.166 09/29 14:34
推 andy2007:是3(2n-1) n用2k代入嗎 我糊塗了140.125.205.230 09/29 15:11
a_n = 3(2n-1)
a_2k = 3(4k-1)
當 n = 2k 也就是n是偶數的情況
a_n = a_2k
這沒有問題吧?
推 andy2007:逆推 3n[2(n+1)-1] 好像怪怪的140.125.205.230 09/29 15:26
請交代一下你的逆推過程
我想知道逆推的方式
推 changefly:A = Σ3[2*2k-1] = 6(n+1)n - 3n 59.124.112.124 09/29 15:29
→ changefly:請問這段是如何整理出來的呢? 59.124.112.124 09/29 15:29
A = a_2+ a_4+ ...+ a_(2n)
= Σa_2k for k = 1 to n
= Σ3(4k-1)
= [12Σk] - 3n
= 6n(n+1) - 3n
P.S. Σk for k = 1 to n
= n(n+1)/2
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◆ From: 122.124.107.17
※ 編輯: Honor1984 來自: 122.124.107.17 (09/30 00:06)
推 andy2007:6(n+1)n - 3n → 3n[2(n+1)-1] 140.125.205.230 09/30 08:00
→ andy2007:我是直接提出公因式140.125.205.230 09/30 08:00
→ andy2007:沒有運用到級數的概念Orz...我錯了140.125.205.230 09/30 08:01
→ andy2007:感謝高手指導 我真是數學天兵 Orz...140.125.205.230 09/30 08:02
推 changefly:原來如此,我忘了基本觀念了,感謝!! 59.124.112.124 09/30 14:25