SECTION 2.1 Elementary Number Theory 105
a(n+ 2)^2 − 2 a(n+ 1)^2 +an^2 = 1,which quickly reduces to 2a = 1, soa=^12. Next, we findb andc by
using the initial conditions:
c = 2
1
2
+b+c = 4.This quickly leads tob=^32 , c= 2 and so the solution is given by
un=^12 n^2 +^32 n+ 2, n= 0, 1 , 2 ,....Exercises
- Let (un)n≥ 0 be an arithmetic sequence. Prove that the sequence
(eun)n≥ 0 is a geometric sequence. - Let (un)n≥ 0 be a geometric sequence withun>0 for alln. Prove
that (logun)n≥ 0 is an arithmetic sequence. - Consider the “counting sequence” 1, 2, 3,....
(a) Represent this sequence as the solution of an inhomogeneous
first-order linear difference equation.
(b) Represent this sequence as the solution of a homogeneous
second-order linear difference equation. Find the general so-
lution of this second-order difference equation.- Solve the linear difference equationun+1=− 2 un, n= 0, 1 , 2 ,...,
whereu 0 = 2 - Solve the second-order difference equationun+2=− 4 un+1+5un, n=
0 , 1 , 2 ,...whereu 0 = 1 =u 1. - Solve the second-order difference equationun+2=− 4 un+1− 4 un, n=
0 , 1 , 2 ,...whereu 0 = 1, u 1 = 0.