Advanced book on Mathematics Olympiad

Real Analysis 463

304.Assume that we have found such numbers for everyn. Thenqn+ 1 (x)−xqn(x)must
be divisible byp(x). But

qn+ 1 (x)−xqn(x)=xn+^1 −an+ 1 x−bn+ 1 −xn+^1 +anx^2 +bnx

=−an+ 1 x−bn+ 1 +an(x^2 − 3 x+ 2 )+ 3 anx− 2 an+bnx

=an(x^2 − 3 x+ 2 )+( 3 an+bn−an+ 1 )x−( 2 an+bn+ 1 ),

and this is divisible byp(x)if and only if 3an+bn−an+ 1 and 2an+bn+ 1 are both
equal to zero. This means that the sequencesanandbnare uniquely determined by the
recurrencesa 1 =3,b 1 =−2,an+ 1 = 3 an+bn,bn+ 1 =− 2 an. The sequences exist and
are uniquely defined by the initial condition.

305.Divide through by the product(n+ 1 )(n+ 2 )(n+ 3 ). The recurrence relation
becomes

xn

n+ 3

= 4

xn− 1

n+ 2

+ 4

xn− 2

n+ 1

.

The sequenceyn=xn/(n+ 3 )satisfies the recurrence

yn= 4 yn− 1 − 4 yn− 2.

Its characteristic equation has the double root 2. Knowing thaty 0 =1 andy 1 =1, we
obtainyn= 2 n−n 2 n−^1. It follows that the answer to the problem is

xn=(n+ 3 ) 2 n−n(n+ 3 ) 2 n−^1.

(D. Bu ̧sneag, I. Maftei,Teme pentru cercurile ̧si concursurile de matematic ̆a(Themes
for mathematics circles and contests), Scrisul Românesc, Craiova)

306.Definec=b/x 1 and consider the matrix

A=

(

0 c

x 1 a

)

.

It is not hard to see that

An=

(

cxn− 1 cxn

xnxn+ 1

)

.

Using the equality detAn=(detA)n, we obtain

c(xn− 1 xn+ 1 −x^2 n)=(−x 1 c)n=(−b)n.

Hencexn^2 −xn+ 1 xn− 1 =(−b)n−^1 x 1 , which does not depend ona.