作业帮If f is the function defined for all k by f(k) = k^5/16,what is f(2k) in terms of f(k)?A.1/8 f(k)B.5/8 f(k)C.2 f(k)D.10 f(k)E.32 f(k)标答是D,但我怎么求都是E另外一道,If an integer n is to be chosen at random from the integers 1 to 96,inc
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If f is the function defined for all k by f(k) = k^5/16,what is f(2k) in terms of f(k)?A.1/8 f(k)B.5/8 f(k)C.2 f(k)D.10 f(k)E.32 f(k)标答是D,但我怎么求都是E另外一道,If an integer n is to be chosen at random from the integers 1 to 96,inc
If f is the function defined for all k by f(k) = k^5/16,what is f(2k) in terms of f(k)?
A.1/8 f(k)
B.5/8 f(k)
C.2 f(k)
D.10 f(k)
E.32 f(k)
标答是D,但我怎么求都是E
另外一道,If an integer n is to be chosen at random from the integers 1 to 96,inclusive,what is the probability that n(n + 1)(n + 2) will be divisible by
A.1/4
B.3/8
C.1/2
D.5/8
E.3/4
标答是D,我可以求出3/8,但是这样就不包含2、4等正解,所以求正确的解法有木有啊!
If f is the function defined for all k by f(k) = k^5/16,what is f(2k) in terms of f(k)?A.1/8 f(k)B.5/8 f(k)C.2 f(k)D.10 f(k)E.32 f(k)标答是D,但我怎么求都是E另外一道,If an integer n is to be chosen at random from the integers 1 to 96,inc
第一题答案显然是E.
第二题,
首先若n是偶数的话,这个数一定是被8整除的.
证明：令n=2k,k为整数.
n(n + 1)(n + 2)=2k(2k+1)(2k+2)= 4k(k+1)(2k+1),
而这里,k(k+1)必为一奇一偶,所以还藏着一个因子2.
此外,n为奇数的时候,令n=2k+1,k为整数.
n(n + 1)(n + 2)=(2k+1)(2k+2)(2k+3)=2(k+1)(2k+1)(2k+3)=2(k+1)n(n+2)
现在已经知道了n为奇数,那么n+2仍为奇数,所以必有k+1被4整除才可以满足.
当n∈{1,2,...96}的时候,使n为奇数的相应的k为0,1,2,...47
相应的,k+1的取值有：1,2,...48
其中4的倍数有：4,8,12...48 一共是48/4=12个
于是当n为奇数的时候满足条件的数有12个.
综上所述,可选的n有：所有的偶数（96/2=48个）以及12个奇数
共计60个数
60/96=5/8
第一题我算的也是E
第二题先试试偶数2,4,6.。。。96.。。都可以。再试试奇数。其中只能（N+1） 被8整除才行。共有（96-8）/8 +1 =12 个。因此概率为（48+12）/96 =5/8...考试中这种题应该很少吧。