The sequence s1, s2, s3, ..., sn, ... is such that ~$s_n=\frac{1}{n}-\frac{1}{n+1}$~for all integers n ≥ 1. If K is a positive integer, is the sum of the first k terms of the sequence greater than ~$\frac{9}{10}$~ ?
(1) k > 10
(2) k < 19
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
题目分析:
前k项的和为~$(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+……+(\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{k+1}=\frac{k}{k+1}$~,即求和过程中中间项的前半部分和前一项的后半部分相互抵消。
要使得前k项的和大于~$\frac{9}{10}$~,则~$\frac{k}{k+1}>\frac{9}{10}$~,解得k>9。
(1) 条件1给出k>10,满足k>9的条件,能够判断前k项的和大于~$\frac{9}{10}$~,题目得解。
(2) 条件2给出k<19,也不能保证k>9,因此也不能判断前k项的和是否大于~$\frac{9}{10}$~。即若9<k<19,则前k项的和大于~$\frac{9}{10}$~,否则前k项的和小于~$\frac{9}{10}$~。条件2不满足要求。
因此,本题答案为(A),仅条件1可以解题,条件2不行。
观察数列求和展开之后的结果!不要急于硬算!
记忆题
老师,怎么得出前k项的和是k÷ (k+1)的,所有数相加再抵消后不是为1-1÷k吗?
否则这道题的答案一定是D!
老师!这道题是不是题干公式的加号应该改成减号?!
你说的没错,改过来了,谢谢!
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