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University of Miami MTH 210230 solution manual 4

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Chapter 4 Solutions An Introduction to Mathematical Thinking: Algebra and Number Systems William J. Gilbert and Scott A. Vanstone, Prentice Hall, 2005 Solutions prepared by William J. Gilbert and Alejandro Morales Exercise 4-1: Calculate the following. 5 3 Solution: 5 3 = 5 3 · · 4 2 ...

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Chapter 4 Solutions
An Introduction to Mathematical Thinking:
Algebra and Number Systems
William J. Gilbert and Scott A. Vanstone, Prentice Hall, 2005

Solutions prepared by William J. Gilbert and Alejandro Morales



Exercise 4-1:  
5
Calculate the following.
3
Solution:

 
5 5·4·3
=
3 3·2·1
60
= = 10.
6


Exercise 4-2:
Calculate
  the following.
10
6
Solution:    
10 10 10 · 9 · 8 · 7
= = = 210
6 4 4·3·2·1


Exercise 4-3:
8!
Calculate the following.
(4!)2
Solution:

 
8! 8
=
(4!)2 4
8·7·6·5
=
4·3·2·1
= 70.


4.1

,Exercise 4-4:
Calculate the following. 100! − 99!.
Solution:


100! − 99! = 99!(100 − 1)
= (98!)(99)2


Exercise 4-5:
   
1 n 1 n−1
Show that = .
n r r r−1
Solution:
For n, r ≥ 1, we have
 
1 n n!
=
n r n · r!(n − r)!
(n − 1)!
= , since n! = n(n − 1)!, n ≥ 1
r!(n − r)!
 
(n − 1)! 1 n−1
= =
r(r − 1)!(n − r)! r r−1

establishing the result.

Exercise4-6:
    
n r n n−s
Show that = .
r s s r−s
Solution:
  
n r n! r!
= ·
r s r!(n − r)! s!(r − s)!
n! (n − s)!
= ·
s!(n − r)!(r − s)! (n − s)!
n! (n − s)!
= ·
s!(n − s)! (r − s)!((n − s) − (r − s))!
  
n n−s
=
s r−s


Exercise
 4-7:
n+2
Find n if = 36.
n
Solution:


4.2

, By definition,  
n+2 (n + 2)!
= .
n 2!n!
Hence, we must solve n2 + 3n + 2 = 72 or n2 + 3n + 70 = 0. By the quadratic
formula, we find n = 7 or n = −10.

Exercise 4-8:
Write the following in sigma notation.
1 3 5 99
+ + + ··· + .
2 4 6 100

Solution:
99
1 3 5 99 X r
+ + + ··· + = .
2 4 6 100 r=1 r + 1


Exercise 4-9:
Write the following in sigma notation.
8 + 15 + 24 + 35 + · · · + (n2 − 1).
Solution:
n
X
8 + 15 + 24 + 35 + · · · + (n2 − 1) = (r2 − 1).
r=3



Exercise 4-10:
Write the following in sigma notation.

ak + a2k + a4k + a8k + a16k + · · · + a256k

Solution:
8
X r
ak + a2k + a4k + a8k + a16k + · · · + a256k = a2 k .
r=0



Exercise 4-11:
Prove, by induction, the following results for all n ∈ P.
12 + 22 + 32 + · · · + n2 = n(n+1)(2n+1)
6 .
Solution:
1(1+1)(2·1+1) 6
(i) When n = 1 the assertion is true since 6 = 6 = 1.

4.3

, (ii) Suppose that
k(k + 1)(2k + 1)
12 + 22 + 32 + · · · + k 2 = .
6
Then

12 + 22 + 32 + · · · + k 2 + (k + 1)2
k(k + 1)(2k + 1) 6(k + 1)(k + 1)
= +
6 6
(k + 1)(k(2k + 1) + 6(k + 1)
=
6
(k + 1)(k(2k + 1) + 2(2k + 1) + 2(k + 2))
=
6
(k + 1)(k + 2)(2k + 1 + 2) (k + 1)(k + 2)(2(k + 1) + 1)
= = .
6 6
Hence the assertion is true for n = k + 1.
Therefore, by the principle of mathematical induction, the assertion is true
for all n ∈ P.

Exercise 4-12:
Prove, by induction, the following results for all n ∈ P.
 2
3 3 3 3 n(n + 1)
1 + 2 + 3 + ··· + n =
2
Solution: h i2
(i) If n = 1, then 13 = 1(1+1)
2 . Hence the proposition is valid for n = 1.
h i2
(ii) Assume that 13 + 23 + · · · + k 3 = k(k+1)
2 for some k ∈ P. Then

13 + 23 + · · · + k 3 + (k + 1)3
 2  2 
k(k + 1) 3 2 k
= + (k + 1) = (k + 1) + (k + 1)
2 4
 2
(k + 2)2

k + 4k + 4
= (k + 1)2 = (k + 1)2
4 22
 2
(k + 1)(k + 2)
= .
2
Thus the proposition is valid for n = k + 1. Therefore, by the principle of
mathematical induction, the proposition is true for all n ∈ P.

Exercise 4-13:
Prove by induction the following results for all n ∈ P.

n(n + 1)(6n3 + 9n2 + n − 1)
14 + 24 + 34 + · · · + n4 =
30

4.4

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