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 ~$\frac{1}{4}$~
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.
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