Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12: Sequences; Induction; the Binomial Theorem
98. Yes, a sequence can be both arithmetic and
geometric. For example, the constant sequence
3,3,3,3,..... can be viewed as an arithmetic
sequence with a 1 = 3 and d = 0. Alternatively,
the same sequence can be viewed as a geometric
sequence with a 1 = 3 and r = 1.
99 – 100. Answers will vary.
101. Answers will vary. Both increase (or decrease)
exponentially, but the domain of a geometric
sequence is the set of natural numbers while the
domain of an exponential function is the set of
all real numbers.
Section 12.4
1. I: n = 1: 2⋅ 1= 2 and1(1 + 1) = 2
II: If 2+ 4+ 6+ + 2 k = k( k+
1) , then
2+ 4+ 6+ + 2k+ 2( k+
1)
= ( 2+ 4+ 6+ + 2k)
+ 2( k+
1)
= kk ( + 1) + 2( k+
1)
2
= k + 3k
+ 2
= ( k + 1)( k + 2)
= ( k+ 1) ⎡⎣( k+ 1)
+ 1⎤⎦
Conditions I and II are satisfied; the statement is
true.
2. I: n = 1: 4⋅1− 3= 1and1(2⋅1− 1) = 1
II: If 1+ 5+ 9 + + (4k − 3) = k(2k−1)
, then
1+ 5+ 9 + + (4k− 3) + [4( k+ 1) −3]
= [ 1+ 5+ 9 + + (4k− 3) ] + 4k+ 4−3
= k(2k− 1) + 4k
+ 1
2
= 2k − k+ 4k+
1
2
= 2k
+ 3k+
1
= ( k+ 1)(2k
+ 1)
= ( k+ 1) ⎡⎣2( k + 1)
−1⎤⎦
Conditions I and II are satisfied; the statement is
true.
3. I:
1
n = 1: 1+ 2 = 3and ⋅ 1(1 + 5) = 3
2
1
II: If 3+ 4+ 5 + + ( k+ 2) = ⋅ k( k+
5) , then
2
3+ 4+ 5 + + ( k+ 2) + [( k+ 1) + 2]
= [ 3+ 4+ 5 + + ( k+ 2) ] + ( k+
3)
1
= ⋅ kk ( + 5) + ( k+
3)
2
1 2 5
= k + k + k+
3
2 2
1 2 7
= k + k + 3
2 2
1 2
= ⋅ ( k + 7k+
6)
2
= 1 ⋅ ( k+ 1)( k + 6)
2
= 1 ⋅ ( k+ 1)[( k + 1) + 5]
2
Conditions I and II are satisfied; the statement is
true.
4. I: n = 1: 2⋅ 1+ 1= 3and1(1 + 2) = 3
II: If 3+ 5+ 7 + + (2k+ 1) = k( k+
2) , then
3 + 5 + 7 + + (2k+ 1) + [2( k+ 1) + 1]
= [ 3+ 5+ 7 + + (2k+ 1) ] + (2k+
3)
= kk ( + 2) + (2k+
3)
2
= k + 2k+ 2k+
3
2
= k + 4k+
3
= ( k + 1)( k + 3)
= ( k + 1)[( k+ 1) + 2]
Conditions I and II are satisfied; the statement is
true.
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