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题库 / Prep2007E1-PS-37
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by ~$\left ( -1 \right )^{k+1}\left ( \frac{1}{2^{k}} \right )$~. If T is the sum of the first 10 terms in the sequence, then T is
greater than 2
between 1 and 2
between ~$\frac{1}{2}$~ and 1
between ~$\frac{1}{4}$~ and ~$\frac{1}{2}$~
less than ~$\frac{1}{4}$~
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等比数列求和公式：Sn=a1*(1-q^n)/(1+q)
首项是1/2，公比是-1/2，项数是10，Sn=1/3(1023/1024)=1/3,选D
a = 1/2 , q= -1/2, n = 10.
等比数列
S=( (1/2) * (1- (-1/2)^10) ) / (1-(-1/2)
S=(1023/1024 ) / 3
Since 1023/1024 is close to 1,
1/3 is between 1/2 and 1/4.
So the answer is D
笔记：这个题竟然没看出来等比数列，绝了。
Sn=a1*(1-q^n)/(1-q)（q≠1)
公比是-1/2
a1=1/2
求和
2^10=1024
算完等于1023/1024 *1/3
近似于1/3
1/3在1/4和1/2之间选D
等比数列公式要牢记！！
方法没有快速反应过来
等比数列的知识
+1 第一次Prep模考一直充满时间紧迫感导致：通项公式里是相乘，算到演算纸上却变成了相加，华丽丽地后面全推导错误
T= (1023/1024)/3
1/4 < 1/3 < 1/2
首项1/2,公比(-1/2)，n=10
s= (1/2)(1-(-1/2)^10)/(1-(-1/2)=1023/1024/3小于1/2大于1/4
解法如下：
the kth term of a certain sequence is given by （-1）^(k+1)乘以(1/2^k)
说明通项公式为：an=(-1)·（-1/2）^k
又因为K=10,已经是一个较大的数了,1-q^n,在n较大的情况下,是可以忽略的,因此
sn=a1/（1-q）,代入即为:1/2除以1-（-1/2）,答案为1/3.
首项1/2,公比(-1/2)，n=10
s= (1/2)(1-(-1/2)^10)/(1-(-1/2)=1023/1024/3约等于1/3的样子
https://gmatclub.com/forum/for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html