MATH 290 – EXAM 3 – SOLUTIONS
1. (16 pts.) Prove by induction that for all natural numbers
n
,
n
X
k
=1
(
k
·
k
!) = (
n
+ 1)!

1.
Solution:
Assume that
n
= 1. Then
1
X
k
=1
(
k
·
k
!) = 1
·
1! = 1 = 2!

1. Hence the result holds
for
n
= 1.
Let
n
be any natural number and assume that
n
X
k
=1
(
k
·
k
!) = (
n
+ 1)!

1. We
must show that
n
+1
X
k
=1
(
k
·
k
!) = (
n
+ 2)!

1. By the induction hypothesis,
n
+1
X
k
=1
(
k
·
k
!) =
n
X
k
=1
(
k
·
k
!) + (
n
+ 1)(
n
+ 1)! = (
n
+ 1)!

1 + (
n
+ 1)(
n
+ 1)!
.
Also,
(
n
+1)!

1+(
n
+1)(
n
+1)! = (
n
+1)![1+(
n
+1)]

1 = (
n
+1)!(
n
+2)

1 = (
n
+2)!

1
.
This is what we wanted to prove.
Therefore by the Principle of Mathematical Induction, for all natural numbers
n
,
n
X
k
=1
(
k
·
k
!) = (
n
+ 1)!

1.
2. (16 pts.) Prove by induction that for all natural numbers
n
≥
4, 2
n
< n
!.
Solution:
Assume that
n
= 4. Then 2
n
= 2
4
= 16
<
24 = 4!. Hence the result holds for
n
= 4.
Let
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 Fall '08
 Alligood,K
 Advanced Math, Natural Numbers, Mathematical Induction, Natural number, Peano axioms, k=1

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