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 ~$\frac{1}{4}$~
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
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
1/4 < 1/3 < 1/2
s= (1/2)(1-(-1/2)^10)/(1-(-1/2)=1023/1024/3小于1/2大于1/4
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.
s= (1/2)(1-(-1/2)^10)/(1-(-1/2)=1023/1024/3约等于1/3的样子