Solved Problems on Binomial Theorem - Quiz 4 | MATH 213, Quizzes of Discrete Mathematics

Material Type: Quiz; Class: Basic Discrete Mathematics; Subject: Mathematics; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;

Typology: Quizzes

Pre 2010

Uploaded on 03/10/2009

koofers-user-8tv
koofers-user-8tv ๐Ÿ‡บ๐Ÿ‡ธ

10 documents

1 / 1

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 213, Section B1, Quiz 4 (Solution); Friday, February 8, 2008
1.
Prove that for every integer nโ‰ฅ1 we have
n
X
j=1 ๎˜’n
j๎˜“5j= 6nโˆ’1.
Solution.
By the Binomial Theorem for every x, y โˆˆRwe have
(x+y)n=
n
X
j=0 ๎˜’n
j๎˜“xnโˆ’jyj.
Substituting x= 1, y= 5 in the above formula, we get:
6n= (1 + 5)n=
n
X
j=0 ๎˜’n
j๎˜“1nโˆ’j5j=
n
X
j=0 ๎˜’n
j๎˜“5j=
=๎˜’n
0๎˜“50+
n
X
j=1 ๎˜’n
j๎˜“5j= 1 +
n
X
j=1 ๎˜’n
j๎˜“5j.
Therefore n
X
j=1 ๎˜’n
j๎˜“5j= 6nโˆ’1,
as required.
1

Partial preview of the text

Download Solved Problems on Binomial Theorem - Quiz 4 | MATH 213 and more Quizzes Discrete Mathematics in PDF only on Docsity!

Math 213, Section B1, Quiz 4 (Solution); Friday, February 8, 2008

Prove that for every integer n โ‰ฅ 1 we have

โˆ‘^ n

j=

n

j

j = 6

n โˆ’ 1.

Solution.

By the Binomial Theorem for every x, y โˆˆ R we have

(x + y)

n

โˆ‘^ n

j=

n

j

x

nโˆ’j y

j .

Substituting x = 1, y = 5 in the above formula, we get:

n = (1 + 5)

n

โˆ‘^ n

j=

n

j

nโˆ’j 5

j

โˆ‘^ n

j=

n

j

j

n

0

0

โˆ‘^ n

j=

n

j

j = 1 +

โˆ‘^ n

j=

n

j

j .

Therefore โˆ‘n

j=

n

j

j = 6

n โˆ’ 1 ,

as required.

1