MACM 101 — Discrete Mathematics I
Exercises on Induction and Combinatorics. Due:
Monday, November 15th (at the beginning of
the class)
Reminder: the work you submit must be your own. Any collaboration and
consulting outside resources must be explicitly mentioned on your submis
sion.
1.
Prove that for every positive integer
n
1
·
2
1
+ 2
·
2
2
+ 3
·
2
3
+
...
+
n
·
2
n
= (
n

1)2
n
+1
+ 2
.
Solution:
We use induction. Let
P
(
n
) denote this equality for the integer
n
.
Basis case.
P
(1) means the equality 1
·
2 =
1(1+1)(1+2)
3
, which is obvi
ously true.
Inductive step.
Suppose that
P
(
k
) is true, that is,
1
·
2 + 2
·
3 + 3
·
4 +
...
+
k
·
(
k
+ 1) =
k
(
k
+ 1)(
k
+ 2)
3
.
We have to prove
P
(
k
+ 1):
1
·
2+2
·
3+3
·
4+
...
+
k
·
(
k
+1)+(
k
+1)
·
(
k
+2) =
(
k
+ 1)(
k
+ 2)(
k
+ 3)
3
.
We have
1
·
2 + 2
·
3 + 3
·
4 +
...
+
k
·
(
k
+ 1) + (
k
+ 1)
·
(
k
+ 2)
=
k
(
k
+ 1)(
k
+ 2)
3
+ (
k
+ 1)
·
(
k
+ 2)
=
k
(
k
+ 1)(
k
+ 2) + 3(
k
+ 1)(
k
+ 2)
3
=
(
k
+ 1)(
k
+ 2)(
k
+ 3)
3
1
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(The gossip problem.)
Suppose there are
n
people in a group,
each aware of a scandal no one else in the group knows about.
These people communicate by telephone; when two people in
the group talk, they share information about all scandals each
knows about. The gossip problem asks for
G
(
n
)
, the minimum
number of telephone calls that are needed for all
n
people to
learn about all the scandals.
Prove that
G
(
n
)
≤
2
n

4
.
Solution:
We prove this statement for all
n
such that
n
≥
4.
Denote by
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 Summer '08
 PEARCE
 Mathematical Induction, Inductive Reasoning, Natural number

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