MATH 111
Exam 2
Solutions
1. (20 pts) Prove the following:
k
X
For all k ∈ N,
(2j − 1) = k 2 .
j=1
Proof : We use induction on k. Note that P (k) is the statement
k
X
(2j − 1) = k 2
j=1
For k = 1: Then
k
1
X
X
(2j − 1) =
(2j − 1) = 2 · 1 − 1 = 1 = 12
j=1
j=1
So P (1) is true.
Now assume that P (n) is true for some n ∈ N. Thus
n
X
(2j − 1) = n2
j=1
We need to show that
n+1
X
(2j − 1) = (n + 1)2
j=1
We have that
n+1
X
(2j − 1) =
j=1
!
n
X
(2j − 1) + (2(n + 1) − 1)
Definition of sum
j=1
2
= n + (2n + 1)
Induction hypothesis
= n2 + 2n + 1
= (n + 1)2
2
So P (n + 1) is true.
k
X
k
2. (10 pts) Use the Binomial Theorem to evaluate
2
where k ∈ Z≥0 .
m
m=0
m
Proof : Let k ∈ Z≥0 . We use the Binomial Theorem to expand (2 + 1)k :
k k k
X
k m k−m X k m X m k
(2 + 1) =
2 ·1
=
2 =
2
m
m
m
m=0
m=0
m=0
k
1
We easily get that (2 + 1)k = 3k . Hence
k
X
k
= 3k
2
m
m=0
2
m
3. (20 pts) Consider the following recurrence relation:
x1 = 0, x2 = 4
xn = 4xn−1 − 4xn−2
if n ≥ 3
n
Prove that xn = (n − 1)2 all n ≥ 1.
Proof : We use strong induction on k to prove the statement P (k):
xk = (k − 1)2k
For k = 1: x1 = 0 and (1 − 1) · 21 = 0 · 2 = 0. So P (1) is true.
For k = 2: x2 = 4 and (2 − 1) · 22 = 1 · 4 = 4. So P (2) is true.
Now assume that P (k) is true for k = 1, 2, . . . , n for some n ≥ 2. So
xk = (k − 1) · 2k
for k = 1, 2, . . . , n
Then
xn+1 =
=
=
=
=
=
=
=
4xn − 4xn−1
4 · (n − 1) · 2n − 4 · ((n − 1) − 1) · 2n−1
4 · (n − 1) · 2n − 4 · (n − 2) · 2n−1
4 · (n − 1) · 2 · 2n−1 − 4 · (n − 2) · 2n−1
4 · 2n−1 ((n − 1) · 2 − (n − 2))
22 · 2n−1 (2n − 2 − n + 2)
2n+1 · n
((n + 1) − 1) · 2n+1
(1)
(2)
(1): We use the recursive definition for xn+1 . Note that we are allowed to use this since
n + 1 ≥ 3. We have to be careful here because the recursive definition does not hold for
x2 . This is one of the reasons we have to consider two base cases.
(2): We use the induction hypothesis for k = n and k = n − 1. Note that we really need
strong induction and not regular induction. This will work only when n − 1 ≥ 1. So
n ≥ 2. This is another reason why we have to consider two base cases.
2
So P (n + 1) is true.
4. (20 pts) Prove that {5m − 6 : m ∈ Z} = {4 − 5m : m ∈ Z}.
Proof : Put A = {5m − 6 : m ∈ Z} and B = {4 − 5m : m ∈ Z}. We prove that A = B
using mutual inclusion.
2
First, we show that A ⊆ B. Let a ∈ A. Then a = 5m − 6 for some m ∈ Z. So
a = 5m − 6 = 4 − 5(2 − m). Since 2 − m ∈ Z, we get that a ∈ B. Thus A ⊆ B.
Next, we show that B ⊆ A. Let b ∈ B. Then b = 4 − 5m for some m ∈ Z. So
b = 4 − 5m = 5(2 − m) − 6. Since 2 − m ∈ Z, we get that b ∈ A. Thus B ⊆ A.
Since A ⊆ B and B ⊆ A, we conclude that A = B.
2
5. (20 pts) Let A, B, C be sets. Prove that (A ∩ B) \ C = (A \ C) ∩ (B \ C).
Proof : We prove this using mutual inclusion.
First, we show that (A ∩ B) \ C ⊆ (A \ C) ∩ (B \ C). Let x ∈ (A ∩ B) \ C. Then x ∈ A ∩ B
and x ∈
/ C. Since x ∈ A ∩ B, we get that x ∈ A and x ∈ B.
Since x ∈ A and x ∈
/ C, we have that x ∈ A \ C.
Since x ∈ B and x ∈
/ C, we have that x ∈ B \ C.
Since x ∈ A \ C and x ∈ B \ C, we get that x ∈ (A \ C) ∩ (B \ C).
Thus (A ∩ B) \ C ⊆ (A \ C) ∩ (B \ C).
Next, we show that (A \ C) ∩ (B \ C) ⊆ (A ∩ B) \ C. Let x ∈ (A \ C) ∩ (B \ C). Then
x ∈ A \ C and x ∈ B \ C.
Since x ∈ A \ C, we have that x ∈ A and x ∈
/ C.
Since x ∈ B \ C, we have that x ∈ B and x ∈
/ C.
Since x ∈ A and x ∈ B, we have that x ∈ A ∩ B.
Since x ∈ A ∩ B and x ∈
/ C, we get that x ∈ (A ∩ B) \ C.
Thus (A \ C) ∩ (B \ C) ⊆ (A ∩ B) \ C.
Since (A ∩ B) \ C ⊆ (A \ C) ∩ (B \ C) and (A \ C) ∩ (B \ C) ⊆ (A ∩ B) \ C, we get that
(A ∩ B) \ C = (A \ C) ∩ (B \ C).
2
6. (10 pts) TRUE or FALSE? If the statement is true, prove it; if the statement is false,
provide a counterexample.
Let A, B ⊆ X. Then (A ∪ B)C = AC ∪ B C .
Solution : FALSE: A correct statement is (A ∪ B)C = AC ∩ B C (this is one of De
Morgan’s Laws).
We provide a counterexample to the original statement.
Let A = {1}, B = {2} and X = {1, 2, 3}. Then A, B ⊆ X and A∪B = {1}∪{2} = {1, 2}.
So
(A ∪ B)C = X \ (A ∪ B) = {1, 2, 3} \ {1, 2} = {3}
3
But AC = X \ A = {1, 2, 3} \ {1} = {2, 3} and B C = X \ B = {1, 2, 3} \ {2} = {1, 3}.
Hence
AC ∪ B C = {2, 3} ∪ {1, 3} = {1, 2, 3}
2
So (A ∪ B)C 6= AC ∪ B C .
Extra Credit (15 pts)
Let p ∈ N. Then for all k ∈ Z≥0 ,
k X
m+p
p
m=0
k+p+1
=
.
p+1
Proof : We prove the statement by induction on k.
For k = 0: Then
k X
m+p
m=0
and
p
=
0 X
m+p
m=0
p
0+p
p
=
=
=1
p
p
0+p+1
p+1
=
=1
p+1
p+1
So P (0) is true.
Now assume that P (n) is true for some n ∈ Z≥0 . Thus
n X
m+p
n+p+1
=
p
p+1
m=0
We need to show that
n+1 X
m+p
m=0
p
n+1+p+1
=
p+1
We will need the following property of binomial coefficients (Corollary 4.20):
k
k
k+1
+
=
for all 1 ≤ m ≤ k
m
m−1
m
Then
! n+1 n X
X
m+p
m+p
n+1+p
=
+
Definition of sum
p
p
p
m=0
m=0
n+p+1
n+1+p
=
+
Induction hypothesis
p+1
p
n+p+1
n+p+1
=
+
p+1
p
n+p+1+1
=
Property of binomial coefficients
p+1
2
So P (n + 1) is true.
4