一次问几个prep上的数学问题，请大家积极参与！

rockedward

1. for every integer k from 1 to 10, inclusive, the kth term for a cewrtain sequence is given by [(-1)^(k+1)]x[1/(2^k)]. 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. less than 1/4

answer: D

求方法，求方法，求方法，除了生硬计算之外。。。
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2. for which of the following function is f(a+b)=f(a)+f(b) for all positive number a and b?

A. f(x)=x^2
B. f(x)=x+1
C. f(x)=√x
D. f(x)=2/x
E. f(x)=-3x

Answer: E

看不懂题目的意思，请大家用通俗易懂的语言解释一下
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3. a certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be form in which at least one member of the group is a senior partner?

A. 48
B. 100
C. 120
D. 288
E. 600

Answer: B

求思路，求思路，求思路。。。
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4. the number of 75 can be written as the squares of 3 different positive integers. What's the sum of these 3 integers?

A. 17
B. 16
C. 15
D. 14
E. 13

Answer: E

感觉这个题不难， 但是怎么也算不出。。。真为自己智商捉急。。求大家解答！

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5. DATA SUFFICIENCY - if p and n are positive integers and p>0, what's the remainder when p^2-n^2 is divided by 15?
    (1) the remainder when p+n divided by 5 is 1.
    (2) the remainder when p-n divided by 3 is 1.

Answer: E - Statement (1) and (2) TOGETHER are NOT sufficient.

求思路， 求思路，求思路。。。。

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感谢大家的热心帮助和踊跃回答！解答请标明题号。

rockedward

顶起顶起顶起

qoo7300293

1.  我的想法是：
第一項：1/2   第二項：-1/4  第二項：1/8....
1/2-1/4+1/8....  第一二項會加會是1/4 第三四項相加是1/4的一半，所以最後應該會是個趨近1/2又大於1/4的式子

2.  把a b a+b 都代進abcde選項中的算式，看哪一個兩邊相等就好了，C選項應該是根號x吧!?  

3.至少一個senior--> 1s2j  2s1j 3s
C4取1*C6取2+C4取2*C6取1+C4取3=100

4.75第於三個數的平方和
75=7^2+5^2+1^2
7+5+1=13

5. p^2-n^2=(p+n)(p-n)
1. p+n=5k+1  ---->不知p-n
2. p-n=3t+1    ---->不知p+n
合併：(p+n)(p-n)=(5k+1)(3t+1)=15kt+3t+5k+1
知道第一個15kt一定整除，但後面的3t,5k由於是任一數，所以不一定整除15

rockedward

qoo7300293 发表于 2015-10-4 11:58
1.  我的想法是：
第一項：1/2   第二項：-1/4  第二項：1/8....
1/2-1/4+1/8....  第一二項會加會是1/4  ...

第二题确实是根号x， 已更改。 感谢大神！