Summation
回答 (5)
2012-04-19 5:35 pm
✔ 最佳答案
答案：13你地攪咩呀，而家 i 係變數喎，n 係常數。
即係每個項都係 n，問題係有幾多個項0者……
由正整數 6 到正整數 n，總共有 n - 6 + 1 咁多個項。
所以個總和就係 (n - 6 + 1) 咁多個 n 囉。
即係
(n - 6 + 1)n = 104
(n - 5)n = 104
n^2 - 5n - 104 = 0
n = 13 (捨去 n = -8)
參考: knowledge
2012-04-19 2:56 am
n n 5
Σn=Σn-Σn=n(n+1)/2 - 5(5+1)/2=104
6 1 1
n^2+n-30=208
n^2+n-238=0
n= -1±√[1-4(1)(-238)] / 2
= -0.5± √953
n= -0.5+√953/2 or -0.5-√953/2
n must be positive
∴n= -0.5+√953/2 =14.93534904...//
Σn=Σn-Σn=n(n+1)/2 - 5(5+1)/2=104
6 1 1
n^2+n-30=208
n^2+n-238=0
n= -1±√[1-4(1)(-238)] / 2
= -0.5± √953
n= -0.5+√953/2 or -0.5-√953/2
n must be positive
∴n= -0.5+√953/2 =14.93534904...//
2012-04-19 1:08 am
(1+n)n/2 - (1+5)5/2 = 104
n+n^2 - 30 = 208
n^2+n-238 = 0
=> n is not an integer.
n+n^2 - 30 = 208
n^2+n-238 = 0
=> n is not an integer.
2012-04-18 6:41 pm
Σn from i = 6 to i = 15 is 105.
2012-04-18 7:55 am
∑ (6->n) n = 104
n(n - 5) = 104
n^2 - 5n - 104 = 0
(n - 13)(n + 8) = 0
n = 13
n(n - 5) = 104
n^2 - 5n - 104 = 0
(n - 13)(n + 8) = 0
n = 13
收錄日期: 2021-04-13 18:38:02
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https://hk.answers.yahoo.com/question/index?qid=20120417000051KK00891