Download MIDTERM EXAM (Solution)

MATH 2056 Discrete Mathematics II, Fall 2015
October 28, 2015
MIDTERM EXAM (Solution)
1. Compute the product.
4
Y
(−1)j
j=0
Solution:
4
Y
(−1)j = (−1)0 · (−1)1 · (−1)2 · (−1)3 · (−1)4 = (−1)0+1+2+3+4 = (−1)10 = 1.
j=0
2. Transform the expression by making the change of variable i = k + 1.
5
X
k(k − 1)
k=0
Solution: When i = k + 1, k = i − 1, hence:
5
X
k(k − 1) =
6
X
(i − 1)(i − 2)
i=1
k=0
3. Prove by mathematical induction that
n+1
X
i · 2i = n · 2n+2 + 2, for all integers n ≥ 0.
i=1
Solution: Let
P (n) ≡
n+1
X
i · 2i = n · 2n+2 + 2, for all integers n ≥ 0.
i=1
Base Case: n = 0.
The left hand side of the equation equals
0+1
X
i · 2i = 1 · 21 = 2.
i=1
The right hand side equals
0 · 20+2 + 2 = 2.
Thus, P (0) is true.
Inductive Step: Assume P (k) is true for some integer k ≥ 0. That is,
k+1
X
i=1
i · 2i = k · 2k+2 + 2
MATH 2056 Discrete Mathematics II, Fall 2015
October 28, 2015
We have to prove that P (k + 1) is true.
(k+1)+1
X
i · 2i =
i=1
k+1
X
i · 2i + (k + 2) · 2k+2
by the properties of the Σ-notation
i=1
= k · 2k+2 + 2 + (k + 2) · 2k+2
k+2
= (2k + 2) · 2
= 2(k + 1)2
= (k + 1)2
k+2
k+3
by the inductive hypothesis
+2
collecting like terms
+2
factoring out 2
+ 2.
by the properties of the exponents
This establishes that P (k + 1) is true.
Conclusion: We conclude, by mathematical induction, that P (n) is true for all integers n ≥ 0.
4. You have two parents, four grandparents, eight great-grandparents, and so forth. If all your ancestors
were distinct, what would be the total number of your ancestors for the past 40 generations (counting
your parents’ generation as number one)? (Hint: Use the formula for the sum of a geometric sequence.)
Solution: Let ai , i ≥ 1 be the number of your ancestors of the ith generation back. Then, we have
a1 = 2, ak = 2ak−1 for all integers k ≥ 2.
By iteration (not given here), we can discover the explicit formula ai = 2i for all integers i ≥ 1. The
total number of all your ancestors is, then
40
X
i=1
ai =
40
X
2i = 2 + 4 + 8 + . . . + 240 = 2(1 + 2 + 4 + . . . + 239 ) = 2 ·
i=1
240 − 1
= 241 − 2.
2−1
Fun fact:
241 − 2 = 2, 199, 023, 255, 550.
That is almost 2.2 trillion people.
5. A sequence a1 , a2 , a3 , . . . is defined by letting a1 = 3 and ak = 7ak−1 for all integers k ≥ 2. Show that
ak = 3 · 7n−1 for all integers n ≥ 1.
Solution: Let P (n) ≡ an = 3 · 7n−1 , for all integers n ≥ 1. We will prove that P (n) is true by
mathematical induction.
Base case: n = 1. a1 = 3 = 3 · 71−1 , therefore P (1) is true.
Inductive step: Let P (k) be true for some integer k ≥ 1. That is, ak = 3 · 7k−1 . We have to prove
that P (k + 1) is true.
ak+1 = 7ak = 7 · 3 · 7k−1 = 3 · 7k = 3 · 7(k+1)−1 ,
which establishes the claim.
Conclusion: We conclude that an = 3 · 7n−1 for all integers n ≥ 1, by mathematical induction.
6. State the well ordering principle for the integers.
Answer: Every non-empty set of integers that is bounded from below has a least element.
MATH 2056 Discrete Mathematics II, Fall 2015
October 28, 2015
7. Compute the first six terms of the sequence
s0 = 1, s1 = 1, sk = sk−1 + 2sk−2 , for all integers k ≥ 2.
Solution:
s0 = 1,
s1 = 1,
s2 = s1 + 2s0 = 1 + 2 · 1 = 3,
s3 = s2 + 2s1 = 3 + 2 · 1 = 5,
s4 = s3 + 2s2 = 5 + 2 · 3 = 11,
s5 = s4 + 2s3 = 11 + 2 · 5 = 21.
8. For the sequence of the Fibonacci numbers, defined as F1 = F2 = 1, Fk = Fk−1 + Fk−2 , for all integers
k ≥ 3, prove that
2
Fk+1
− Fk2 = Fk−1 Fk+2 , for all integers k ≥ 2.
Solution: We will prove the given property of the Fibonacci numbers by mathematical induction. Let
2
P (n) ≡ Fn+1
− Fn2 = Fn−1 Fn+2 , for all integers n ≥ 2.
Base case: n = 2.
The left hand side equals F32 − F22 = 22 − 12 = 3.
The right hand side equals F1 F4 = 1 · 3 = 3.
Therefore, P (2) is true.
2
Inductive step: Assume P (k) is true for some integer k ≥ 2. Namely, that Fk+1
− Fk2 = Fk−1 Fk+2 .
To prove P (k + 1), consider the following
2
2
2
Fk+2
− Fk+1
= Fk+2
− (Fk2 + Fk−1 Fk+2 )
=
2
Fk+2
− Fk+2 Fk−1 −
by the inductive hypothesis
Fk2
= Fk+2 (Fk+2 − Fk−1 ) −
by the distributive and associative laws
Fk2
= Fk+2 (Fk+1 + Fk − Fk−1 ) −
factoring out Fk+2
Fk2
by the Fibonacci recurrence for Fk+2
= Fk+2 (Fk + Fk−1 + Fk − Fk−1 ) −
= 2Fk Fk+2 −
Fk2
by the Fibonacci recurrence for Fk+1
Fk2
simplification
= Fk (2Fk+2 − Fk )
= Fk (Fk+2 + Fk+1 + Fk − Fk )
factoring out Fk
by the Fibonacci recurrence for Fk+2
= Fk (Fk+2 + Fk+1 )
= Fk Fk+3 .
simplification
by the Fibonacci recurrence for Fk+3
The above is the right hand side in P (k + 1), hence we have proven P (k + 1) is true.
Conclusion: We conclude, by mathematical induction, that
2
Fn+1
− Fn2 = Fn−1 Fn+2 , for all integers n ≥ 2.
Note: We do not actually need to use the inductive hypothesis, a much more straightforward proof is
based on the recurrence only (i.e., the induction is hidden inside the recurrence), as follows
2
2
Fk+2
− Fk+1
= (Fk+2 + Fk+1 )(Fk+2 − Fk+1 )
= Fk+3 (Fk+1 + Fk − Fk+1 )
= Fk+3 Fk .
a2 − b2 = (a + b)(a − b)
by the Fibonacci recurrences for Fk+3 and Fk+2
simplification
MATH 2056 Discrete Mathematics II, Fall 2015
October 28, 2015
9. Use iteration to discover explicit formula for the sequence.
bk−1
, for all integers k ≥ 1.
1 + bk−1
b0 = 1, bk =
Solution:
b0 = 1,
b0
1
1
=
= ,
1 + b0
1+1
2
1
1
b1
b2 =
= 2 1 = ,
1 + b1
3
1+ 2
b1 =
b3 =
1
b2
= 3
1 + b2
1+
1
3
b4 =
1
b3
= 4
1 + b3
1+
1
4
b5 =
1
b4
= 5
1 + b4
1+
1
5
=
1
,
4
=
1
,
5
=
1
, etc.
6
Based on the above,
bn =
1
for all integers n ≥ 0.
n+1
10. Find an explicit formula for the sequence
a0 = 1, a1 = 2, ak = 2ak−1 + 3ak−2 , for all integers k ≥ 2.
Solution: The characteristic equation is
x2 = 2x + 3 ⇔ x2 − 2x − 3 = 0 ⇔ (x − 3)(x + 1) = 0,
which has roots α = 3 and β = −1. Therefore, the general form of the term is
an = A · 3n + B · (−1)n , n ≥ 0.
Form the initial conditions, we have
A+ B =1
3A − B = 2
Adding the two equations, we have
4A = 3 ⇒ A =
3
1
⇒B= .
4
4
Thus,
an =
3 n 1
3n+1 + (−1)n
· 3 + · (−1)n =
, for all integers n ≥ 0.
4
4
4