Download Consider an urn with n red balls

Proof Assignment #2
1.(3 points) Prove the following binomial identity:
2n
n
Pn
=
j=0
n 2
.
j
Hint: Consider an urn with n red balls and n blue balls inside. Show that each side of
the equation equals the number of ways to choose n balls from the urn.
Consider an urn with n red balls and n blue balls inside (so it contains a total of 2n
balls). The number
of ways that we could draw any subset of n balls from the urn (ignoring
the colors) is 2n
. This appears on the lefthand side of our equation.
n
On the other hand, we could think of having two separate urns, one containing the n
red balls and the other containing the n blue balls, and we could count the number of ways
that we could draw a subset of n balls (total) out of the two urns. In other words, we are
counting all of the ways of drawing j red balls and n − j blue balls for values of j ranging
from 0, ..., n.
To choose j red balls and n − j blue balls, we have
n
we showed in class that nj = n−j
. Thus, we have
n
j
·
n
n−j
possibilities. Recall that
2
n
n
n
n
n
·
=
·
=
j
n−j
j
j
j
possible ways of drawing j red balls and n − j blue balls. Therefore, the total number of
ways that we could choose any subset of n balls from the two urns, which contain a total of
2n balls, is
2 X
2 2
n 2
n
n
n
n
.
=
+ ··· +
+
j
n
1
0
j=0
This appears on the righthand side of our equation.
Since the expressions on the lefthand and righthand sides of the equation are counting
the same thing (i.e. the number of ways of drawing any subset of n balls out of a total of
2n) then the two sides must be equal.
2.(2 points) Find two sequences of positive real numbers, {an } and {bn }, such that an ∼ bn
but ann bnn .
Answers will vary for this question. One possible example:
Let an =
1+n
,
n
bn = 1. Then
lim
n→∞
so
n+1
n
1+n
n
1
1+n
= 1,
n→∞
n
= lim
∼ 1.
1
Proof Assignment #2
On the other hand, ann = ( 1+n
)n , bnn = 1n = 1 and
n
( 1+n )n
lim n
= lim
n→∞
n→∞
1
1+n
n
n
=e
(from elementary calculus). Thus, ann bnn .
Bonus (+2 points) Prove:
just one line.
2n
n
< 4n . Do NOT use Stirling’s formula. Your proof should be
From the Binomial Theorem, we know that
2n X
2n
j=0
j
= (1 + 1)2n = 22n = 4n .
P2n
Since 2n
is
only
one
of
the
middle
terms
in
the
sum,
it
is
strictly
less
than
j=0
n
it is certainly the case that
2n
< 4n .
n
2
2n
n
. Thus,