∑6/n*(n+1)*(n+2}= 3/2
答案是3/2,n*(n+1)*(n+2好像可以變成分數互消但我不會,求解釋
數學∑問題高中...
2013-04-01 5:50 am
回答 (3)
2013-04-01 8:37 am
✔ 最佳答案
利用2/n*(n+1)*(n+2) = 1/n(n+1) - 1/(n+1)(n+2) Sn = Σ6/n*(n+1)*(n+2)
= 3 Σ 2/n*(n+1)*(n+2)
= 3 [ 1/n(n+1) - 1/(n+1)(n+2) ]
= 3[ (1/1*2- 1/2*3)+(1/2*3-1/3*4)+......( 1/n(n+1) - 1/(n+1)(n+2) )]
分項對消
= 3 [ 1/1*2 -1/(n+1)(n+2) ]
當n趨近∞時，1/(n+1)(n+2)趨近0
lim(n趨近∞) Sn = 3 [ 1/1*2 -0 ] = 3/2
2013-04-01 00:41:29 補充：
第三行漏了"Σ"
參考: 如露亦如電
2013-04-01 8:32 am
假設 n=1...∞
n=∞
Σ 6/[n(n+1)(n+2)]
n=1
n=∞
＝ Σ [3/n(n+1) - 3/(n+1)(n+2)]......分項對消
n=1
n=∞
＝ Σ [3/n - 3/(n+1) - 3/(n+1) + 3/(n+2)]......分項對消
n=1
n=∞ n=∞
＝ Σ [3/n - 3/(n+1)] - Σ [3/(n+1) - 3/(n+2)]
n=1 n=1
＝ 3 - 3/2＝ 3/2
n=∞
Σ 6/[n(n+1)(n+2)]
n=1
n=∞
＝ Σ [3/n(n+1) - 3/(n+1)(n+2)]......分項對消
n=1
n=∞
＝ Σ [3/n - 3/(n+1) - 3/(n+1) + 3/(n+2)]......分項對消
n=1
n=∞ n=∞
＝ Σ [3/n - 3/(n+1)] - Σ [3/(n+1) - 3/(n+2)]
n=1 n=1
＝ 3 - 3/2＝ 3/2
2013-04-01 7:24 am
∑從k=1到多少??
這題會用「分項對消」喔
這題會用「分項對消」喔
收錄日期: 2021-05-01 18:41:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130331000015KK05050