Fall 2009
CS131 – Combinatorial Structures
Homework 6
Homework 6, due Oct 27
You must prove your answer to every question.
Do not rely only on the homework for exercise: there are several selfcheck ex
ercises of the easier kind in the book, try to solve them, too!
Problem 1.
(10pts) Prove the following identity:
1

2
·
2
+
3
·
4

4
·
8
+···±
n
·
2
n

1
=
(1

(

2)
n
(3
n
+
1))/9.
(The signs are alternating, this determines
±
.) [Hint: use mathematical induction.]
Solution.
Base case:
n
=
1, then we have just the ﬁrst term, 1. The righthand side
is (1

(

2)
·
4)/9
=
(1
+
8)/9.
Induction step: assume that the statement holds for
n
, we prove it for
n
+
1. We
add (
n
+
1)(

2)
n
to both sides. Then the lefthand side becomes the sum for
n
+
1.
The righthand side becomes
1

(

2)
n
(3
n
+
1)
9
+
(
n
+
1)(

2)
n
=
1

(

2)
n
(3
n
+
1)
+
9(
n
+
1)(

2)
n
9
.
The numerator is
1
+
(

2)
n
(6
n
+
8)
=
1

(

2)
n
+
1
(3
n
+
4)
=
1

(

2)
n
+
1
(3(
n
+
1)
+
1),
which is the numerator of the righhand side for
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 Fall '11
 FredPhelps
 Math, Mathematical Induction, Natural number, Prime number, Divisor, combinatorial structures

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