COT310001, Spring 2000
Assigned: 4/12/2000
S. Lang
Solution Key to Assignment #5 (40 pts.)
Due: 4/19 in class
1.
Use induction on
n
≥
0 to prove the following summation identity:
.
0
2
)
2
(
)
1
(
2
)
1
(
∑
=
+

=

n
i
n
n
n
i
i
We use induction on
n
≥
0.
(Basis Step) Consider
n
= 0.
In this case,
.
0
2
0
2
)
0
2
0
(
0
)
1
(
RHS
The
.
0
0
)
1
(
)
1
(
LHS
The
2
0
0
0
2
=
=
+

=
=

=
∑ 
=
=
i
i
i
Thus, LHS = RHS, so the Basis Step is proved.
(Induction Hypothesis) Consider
n
=
k
.
Suppose
.
0
some
for
0
2
)
2
(
)
1
(
2
)
1
(
≥
∑
=
+

=

k
k
i
k
k
k
i
i
(Induction Step) Consider
n
=
k
+ 1.
We need to prove
(1).
of
RHS
2
)
2
3
(
1
)
1
(
2
)
2
4
2
2
(
1
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
since
,
2
)
1
(
2
)
1
(
2
)
2
)(
1
(
1
)
1
(
Hypothesis
Induction
by the
,
)
1
(
)
1
(
2
)
2
(
)
1
(
summation
of
definition
by the
,
)
1
(
)
1
(
)
1
(
(1)
of
LHS
The
2
2
1
2
2
1
2
1
2
1
0
2
=
+
+
+

=
+
+
+


+

=


=

=

+

+
+

+

=
+

+
+

=
+

+
∑ 
=
+
+
+
+
+
=
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
i
k
k
k
k
k
k
k
i
i
Thus, the Induction Step is proved.
By induction, we proved that
2.
Suppose a function
f
(
n
) is defined for integer
n
≥
0 recursively by the following recurrence:
f
(0) = 7;
f
(1) = 14; and
f
(
n
) =
f
(
n
– 1) + 12
f
(
n
– 2) for
n
≥
2.
Prove that
f
(
n
) = 5(4)
n
+ 2(–3)
n
for all
n
≥
0.
We use strong induction on
n
≥
0.
(Basis Step) We first consider
n
= 0.
In this case, applying the formula for
f
(
n
) yields
f
(0) = 5(4)
0
+ 2(–3)
0
= 5 + 2 = 7, which equals the given initial value for
f
(0).
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 Spring '09
 Mathematical Induction, Recursion, Inductive Reasoning, 0 k, Structural induction

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