Sample Exam Questions
1. Write an inductive proof of the following:
13 + 33 + 53 + … + (2n+1)3 = (n+1)2(2n2+4n+1)
basis step: n=1
13 + ((2 * 1) + 1 )3 = (1 + 1)2 ((2 * 12) + (4 * 1) + 1)
1 + 27 = 4 * 7
28 = 28
inductive step: prove
13 + 33 + 53 + … + (2n+1)3 + (2(n+1) + 1) 3 = (n+1+1)2(2(n+1)2+4(n+1)+1)
substituting:
(n+1)2(2n2+4n+1) + (2(n+1) + 1) 3 = (n+1+1)2(2(n+1)2+4(n+1)+1)
(n+1)2(2n2+4n+1) + (2n+3) 3 = (n+2)2((2n2+4n + 2) +(4n+4)+1)
(n2 + 2n + 1)(2n2+4n+1) + (2n+3) 3 = (n+2)2(2n2+8n + 7)
2n4 + 8n3 + 11n2 + 6n + 1 + (2n+3) 3 = (n2 + 4n + 4)(2n2+8n + 7)
2n4 + 8n3 + 11n2 + 6n + 1 + 8n3 + 36n2 + 54n + 27 = 2n4 + 16n3 + 47n2 + 60n + 28
2n4 + 16n3 + 47n2 + 60n + 28 = 2n4 + 16n3 + 47n2 + 60n + 28
2. How many license plates can be made using either two letters followed by four digits
or two digits followed by four letters?
If duplicates are allowed, then 2 letters/4 digits is: 26*26*104and 4 digits/2 letters
is 264 * 10 * 10; if no duplicates are allowed, it's 26*25*10*9*8*7 and
26*25*24*23*10*9
3. There are 51 houses on a street. Each house has an address between 1000 and 1099,
inclusive. Shat that at least two houses have addresses that are consecutive integers.
In the set {0..99} there are 50 odd numbers and 50 even numbers; to get a subset
of 51 numbers from this set, we can choose as many as 50 evens or 50 odds, but
we must choose at least one from the other subset – so at least 2 addresses must
be consecutive.
4. Suppose 100 babies are born every day of every year. What is the probability that a
particular baby was born on February 29?
A baby has a ¼ chance of being born in a leap year, and a 1/366 chance of being
born on February 29 of that year; so the probability of any baby being born on
leap day is 1/(4*366).
5. Use two of the following methods of proof: direct proof, indirect proof, and proof by
contradiction to prove the following proposition: If n is even, then 3n is even.
direct proof: if n is even, then n=2k and 3n = 6k, which is 2(3k) and therefore
even
indirect proof: if 3n is odd, then 3n = 2k + 1 and n must be (2k+1)/3, which is
also an odd number, since it is the product of odd numbers
proof by contradiction: suppose n is even and 3n is odd; subtracting an even
from an odd should yield an odd, but 3n-n = 2n, which is even; so 3n is off is a
contradiction if n is even.
6. Give a recursive definition of:
a) the set of odd positive integers
a1 = 1; an+1 + an + 2
b) the set of positive integers that are multiples of 5 a1 = 5; a n+1 + 5an
c) the set of positive integer powers of 3
a0 = 1; a n+1 = 3an
7. What is the x4y6 term of (2x + y)10?
C(10,4)(2)4 = (10!/6!4!)(16) = ((10*9*8*7)/(4*3*2))(16) = 5*3*4*7*16