Chapter 12 Sequences; Induction; the Binomial Theorem

Section 12.4: Mathematical Induction
21. I:
2
n = 1: 1 − 1+ 2 = 2 is divisible by 2
2
II: If k − k + 2 is divisible by 2 , then
2 2
( k + 1) − ( k+ 1) + 2= k + 2k+ 1−k− 1+
2
2
= ( k − k+ 2) + (2 k)
2
Since k − k + 2 is divisible by 2 and 2k is
2
divisible by 2, then ( k + 1) − ( k + 1) + 2 is
divisible by 2.
Conditions I and II are satisfied; the statement is
true.
22. I: n = 1: 1(1 + 1)(1 + 2) = 6 is divisible by 6
23. I:
24. I:
II: If kk ( + 1)( k+ 2) is divisible by 6 , then
( k + 1)( k+ 1+ 1)( k+ 1+
2)
= ( k + 1)( k+ 2)( k+
3)
= kk ( + 1)( k+ 2) + 3( k+ 1)( k+
2).
Now, kk ( + 1)( k+
2) is divisible by 6;
and since either k + 1 or k + 2 is even,
3( k + 1)( k + 2)
is divisible by 6.
Thus, ( k + 1)( k+ 2)( k+
3)
= k k+ 1 k+ 2 + 3 k+ 1 k+
2
( )( ) ( )( )
is divisible by 6.
Conditions I and II are satisfied; the statement is
true.
n= 1: If x > 1then x = x > 1.
II: Assume, for some natural number k, that if
k
x > 1 , then x > 1 .
k + 1
Then x > 1, for x > 1,
k+
1 k
x = x ⋅ x > 1⋅ x = x > 1
↑
k
( x > 1)
Conditions I and II are satisfied; the statement is
true.
n= 1: If 0 < x< 1then 0 < x < 1.
II: Assume, for some natural number k, that if
k
0< x < 1, then 0< x < 1.
Then, for 0 < x < 1,
k+
1 k
0< x = x ⋅ x< 1⋅ x = x<
1
k + 1
Thus, 0 < x < 1.
Conditions I and II are satisfied; the statement is
true.
1
1
25. I:
1 1
n= 1: a−bis a factor of a − b = a−
b.
II: If a− b is a factor of a − b , show that
a− b is a factor of a − b .
k+ 1 k+
1 k k
a − b = a⋅a −b⋅b
= aa ⋅ −ab ⋅ + ab ⋅ −bb
⋅
k k k
= a a − b + b ( a−b)
k
( )
k
k+ 1 k+
1
k k k k
Since a− b is a factor of a − b and a−
b
is a factor of a− b, then a− b is a factor of
a
k+ 1 k+
1
− b .
Conditions I and II are satisfied; the statement is
true.
26. I: n = 1:
21 ⋅+ 1 21 ⋅+ 1 3 3
a+ bis a factor of a + b = a + b .
⎡ 3 3 2 2
a + b = ( a+ b)( a − ab+
b )
⎤
⎣
⎦
II:
k
k
2k+ 1 2k+
1
If a+ b is a factor of a + b ,
show that a+ b is a factor of
2( k+ 1) + 1 2( k+ 1)
+ 1
a + b .
2( k+ 1) + 1 2( k+ 1) + 1 2k+ 3 2k+
3
a + b = a + b
= a ⋅ a + a ⋅b −a ⋅ b + b ⋅b
2 2k+ 1 2k+ 1 2k+
1 2 2
= a a + b −b ( a −b
)
2 2k+ 1 2 2k+ 1 2 2k+ 1 2 2k+
1
( )
2k+ 1 2k+
1
Since a+ b is a factor of a + b and
2 2
a+ b is a factor of a − b [( a− b)( a+ b)
] ,
2k+ 3 2k+
3
then a+ b is a factor of a + b .
Conditions I and II are satisfied; the statement is
true.
27. I: 1
n = : ( ) 1
1+ a = 1+ a ≥ 1+ 1⋅
a
II: Assume that there is an integer k for which
the inequality holds. We need to show that if
k
1+ a ≥ 1+ ka then
( )
k + 1
( 1+ a) ≥ 1+ ( k+ 1)
a.
k+
1
k
( 1+ a) = ( 1+ a) ( 1+
a)
≥ ( 1+ ka)( 1+
a)
2
= 1+ ka + a + ka
= 1+ k+ 1 a+
ka
( )
( k )
≥ 1+ + 1 a
Conditions I and II are satisfied, the statement is
true.
2
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