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标题: 求助戰友們幫忙pp37 [打印本页]

作者: hijenny 时间: 2010-7-8 17:33
标题: 求助戰友們幫忙pp37
37.5000-!-item-!-187;#058&004365

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by
. (在檔案裡) If T is the sum of the first 10 terms in the sequence, then T is

(A) greater than 2

(B) between 1 and 2

(C) between 1/2 and 1

(D) between 1/4 and 1/2

(E) between 1/4

我的解法
k代1=1/2
k代2=-1/4
k代3=8/1
類推
得1/2-1/4+1/8-1/16+1/32.........+1/512-1/1024
在一個個通分
才得到正確答案
花了五分鐘= =
我嚇死了
有更快的解法嗎?

作者: hijenny 时间: 2010-7-8 17:34
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by
(在檔案裡). If T is the sum of the first 10 terms in the sequence, then T is

題目

作者: raymondsino 时间: 2010-7-8 17:46

37.5000-!-item-!-187;#058&004365

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by
. (在檔案裡) If T is the sum of the first 10 terms in the sequence, then T is

(A) greater than 2

(B) between 1 and 2

(C) between 1/2 and 1

(D) between 1/4 and 1/2

(E) between 1/4

我的解法
k代1=1/2
k代2=-1/4
k代3=8/1
類推
得1/2-1/4+1/8-1/16+1/32.........+1/512-1/1024
在一個個通分
才得到正確答案
花了五分鐘= =
我嚇死了
有更快的解法嗎?

-- by 会员 hijenny (2010/7/8 17:33:41)

This is a geometric sequence, which q=(-1/2), a1=1/2, the answer should be Sn, when n=10

作者: hijenny 时间: 2010-7-8 18:09
不好意思小妹不才
能再解釋的清楚一點嗎?

作者: hijenny 时间: 2010-7-8 18:09
i appreciate your help.

作者: raymondsino 时间: 2010-7-8 19:26

i appreciate your help.

-- by 会员 hijenny (2010/7/8 18:09:50)

这就是一个等比数列，前10项求和，比为-1/2，首项为1/2

作者: hijenny 时间: 2010-7-8 19:48
我不會用這個
能不能教一下怎麼用
我記得好像有公式

作者: hijenny 时间: 2010-7-8 19:55
鋼查了一下
是
首項/(1-公比)
請問公比怎麼求?

剛把後剪前試了一下還是求不出來

作者: hijenny 时间: 2010-7-8 19:55
多謝大牛指點

作者: raymondsino 时间: 2010-7-8 20:00

多謝大牛指點

-- by 会员 hijenny (2010/7/8 19:55:50)

公比就是每一个后项除以前项的商，为-1/2

计算公式是 a1(1-q^n)/(1-q), a1为首项，q为公比，n是项数

作者: hijenny 时间: 2010-7-8 20:49
謝謝
請問你有其他解法嗎
我很怕被錯公式

作者: chandleryeah 时间: 2010-7-8 21:00
最烦老外拿看不懂的话来绕弱智的题，显摆他们能似的

作者: hijenny 时间: 2010-7-8 23:07
哈哈

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