MATH 31102
Notes
Introduction to Higher Math
Even when we don’t have an explicit formula for a recurrence, we can often work out valuable
information about it using inductive methods:
Proposition 1.
Let
b
n
be given by a recurrence relation:
b
1
= 1
,
b
2
= 2
, and
b
n
=
nb
n

1
+
b
n

2
for
n >
2
. For positive integers
n
,
2

b
n
if and only if
2

n
.
Proof.
Let
P
(
n
) be the assertion that 2

b
n
if and only if 2

n
; alternatively, it could be phrased as
the assertion that
b
n
and
n
have the same parity. The base cases
P
(1) and
P
(2) are easily veriﬁed:
P
(1) is odd as is 1, and
P
(2) is even as is 2.
Now let us ﬁx a
k
≥
2, and inductively assume that
P
(1),
P
(2), .
. . ,
P
(
k
) are all true. Speciﬁcally,
we will use the facts that
P
(
k
) and
P
(
k

1) are true below. Now we have two cases: either 2

(
k
+1),
in which case our goal is to show that 2

b
k
+1
, or 2

(
k
+ 1), in which case our goal is to show that
2

b
k
+1
. We address these casewise:
Case I:
2

k
+ 1
(a.k.a.
k
+ 1
is even).
Since
k
+ 1 is even,
k
is odd and
k

1 is even, and
since we presumed
P
(
k
) and
P
(
k

1) to be true, it thus respectively follows that
b
k
is odd and
b
k

1
is even. Then, we note that
b
k
+1
= (
k
+ 1)
b
k
+
b
k

1
; this is a sum of a product of an even and odd
number with an even number; the result of such a computation is even, so 2
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 Spring '08
 Staff
 Math, Mathematical Induction, Recursion, Natural number, 2 K, 2k

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