幾題數學題目~~急!!

[匿名]

2012-08-11 9:24 am

1.設等差數列<aˇn>，前n向的和為Sˇn，若Sˇn=10，Sˇ2n=17，則Sˇ3n=?

2.從200到1000之間所有7的倍數總和是?

麻煩附詳細計算過程~~急!!!!

回答 (1)

胡雪8°

2012-08-11 11:05 am

✔ 最佳答案

1.
設首項為 a，公比為 d。

Sn :
n[2a + (n - 1)d] / 2 = 10
2na + n²d - nd = 20 ...... [1]

S2n :
(2n)[2a + (2n - 1)d] / 2 = 17
2na + 2n²d - nd = 17 ...... [2]

[2] - [1] :
n²d = -3

S3n
= 3n[2a + (3n - 1)d] / 2
= 3n[2a + (n - 1 + 2n)d] / 2
= 3n[2a + (n - 1)d + 2nd] / 2
= 3n[2a + (n - 1)d + 2nd] / 2
= {3n[2a + (n - 1)d] / 2} + 3n²d
= 3Sn + 3n²d
= 3(10) + 3(-3)
= 21

=====
2.
所求的總和
= 203 + 210 + 217 + ...... + 994

這是等差數列的和。
首項 a = 203
末項 l = 994
項數 n = [(994 - 203)/7] + 1 = 114

所求的總和
= n(a + I) / 2
= 114(203 + 994) / 2
= 68229

參考: 胡雪

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