等差等比求和 小小問題

2013-09-23 1:16 am
考慮等差數列5.5,7,8.5,10,..., 其通項是T(n)

(a)求首n項之和S(n)

我計到S(n)=20

但係個答案係0.75n^2+4.75n

唔知點得出黎

回答 (1)

2013-09-23 1:30 am
✔ 最佳答案
首項 = a = 5.5

公差 = d = 7-5.5 = 1.5

第n項 = T(n)
= a + (n-1)d
= 5.5 + (n-1)(1.5)
= 1.5n + 4

首n項之和 = S(n)
= (n/2)[2a+(n-1)d]
= (n/2)[2*5.5+(n-1)*1.5]
= (n/2)(11+1.5n-1.5)
= (n/2)(1.5n+9.5)
= n(0.75n+4.75)
= 0.75n² + 4.75n


2013-09-22 17:33:16 補充:
或者我解釋一下如何得出 S(n) 的公式。

  S(n) = T(1)+ T(2) +...+T(n)
+ S(n) = T(n)+T(n-1)+...+T(1)
-----------------------------
2 S(n) = n*〔T(1)+T(n)〕

原來 T(1)+T(n) = T(2) + T(n-1) = T(3) + T(n-2) = ...

2 S(n) = n*〔a+a+(n-1)d〕

S(n) = n*〔2a+(n-1)d〕/2

2013-09-22 17:34:38 補充:
數字例子:

S =   1 +  2 +  3 + ... + 100
S = 100 + 99 + 98 + ... +   1
-----------------------------
2S= 101 +101 +101 + ... +101

2S= 101*100

S = 101*50 = 5050


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