MATH 74 HOMEWORK 5 (DUE WEDNESDAY OCTOBER 3)
1.
For
n
∈
N
consider the sequences
A
n
=
∑
n
j
=1
j
2
j
and
B
n
= (
n

1)2
n
+1
+ 2.
We have
A
1
= 12
1
= 2 = 0
·
2
2
+ 2 =
B
1
and for any
n
∈
N
we have
A
n
+1

A
n
=
n
+1
j
=1
j
2
j

n
j
=1
j
2
j
= (
n
+ 1)2
n
+1
and
B
n
+1

B
n
= (
n
2
n
+2
+ 2)

((
n

1)2
n
+1
+ 2)
=
n
2
n
+2

n
2
n
+1
+ 2
n
+1
= 2
n
2
n
+1

n
2
n
+1
+ 2
n
+1
=
n
2
n
+1
+ 2
n
+1
= (
n
+ 1)2
n
+1
.
Thus
A
n
+1

A
n
=
B
n
+1

B
n
for all
n
∈
N
. We conclude that
A
n
=
B
n
for all
n
∈
N
by the sequence equality theorem.
2.
For
n
∈
N
let
P
(
n
) denote the statement
f
(
n
)
(
x
) =
ne
x
+
xe
x
.
P
(1) is the statement that
f
(
x
) =
e
x
+
xe
x
, which is true by the product rule. If
P
(
k
) is true for some
k
∈
N
, that is, if
f
(
k
)
(
x
) =
ke
x
+
xe
x
,
we have
f
(
k
+1
(
x
) =
d
dx
f
(
k
)
(
x
)
by definition
=
d
dx
(
ke
x
+
xe
x
)
by hypothesis
=
ke
x
+ (
e
x
+
xe
x
)
by the sum and product rule
= (
k
+ 1)
e
x
+
xe
x
so that
P
(
k
+ 1) is also true.
By the induction principle we conclude that
P
(
n
)
holds for all
n
∈
N
.
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MATH 74 HOMEWORK 5 (DUE WEDNESDAY OCTOBER 3)
3.
For
n
∈
N
let
P
(
n
) denote the statement that
∑
n
j
=1
a
j
≥
na
n
.
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 Fall '07
 COURTNEY
 Math, Product Rule, Induction principle

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