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{SECT 0 {EXCHG {PARA 262 "" 0 "" {TEXT 265 42 "Inhomogeneous Linear Re
currence Equations:" }}{PARA 262 "" 0 "" {TEXT 266 35 "Method of Undet
ermined Coefficients" }}{PARA 257 "" 0 "" {TEXT 267 0 "" }}{PARA 263 "
" 0 "" {TEXT 268 18 "Date: Nov 13, 2001" }}{PARA 263 "" 0 "" {TEXT 
269 27 "Last Revision: Nov 15, 2001" }}{PARA 257 "" 0 "" {TEXT 270 0 "
" }}{PARA 263 "" 0 "" {TEXT 271 16 "Bent E. Petersen" }}{PARA 264 "" 
0 "" {TEXT 272 17 "bent@alum.mit.edu" }}{PARA 264 "" 0 "" {TEXT 273 
22 "petersen@math.orst.edu" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 
261 "" 0 "" {TEXT -1 32 "Course: Mth 355 (a.k.a. Mth 399)" }}{PARA 
261 "" 0 "" {TEXT -1 15 "Term: Fall 2001" }}{PARA 261 "" 0 "" {TEXT 
-1 12 "File name:  " }{TEXT 274 19 "355f2001_inhrec.mws" }}{PARA 261 "
" 0 "" {TEXT 275 49 "Assignment 7 (problems at end) - due Nov 21, 2001
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 12 "Intr
oduction" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
94 "Maple has no difficulty solving simple inhomogeneous linear recurr
ence equations. For example," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "eqn0:=a(n)=4*a(n-1)-4*a(n-2)
+3^n+4^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn0G/-%\"aG6#%\"nG,*-
F'6#,&F)\"\"\"F.!\"\"\"\"%*&F0F.-F'6#,&F)F.\"\"#F/F.F/)\"\"$F)F.)F0F)F
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "init0:=a(0)=A,a(1)=B;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&init0G6$/-%\"aG6#\"\"!%\"AG/-F(
6#\"\"\"%\"BG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "soln0:=rso
lve(\{eqn0,init0\},a(n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&soln0G
,.*(,&%\"AG!\"\"*&#\"\"\"\"\"#F,%\"BGF,F,F,,&%\"nGF,F,F,F,)F-F0F,F,*&,
&F.F+*&F-F,F(F,F)F,F1F,F)*&,&F0#!#<F-#\"#<F-F)F,F1F,F,*&#\"\"*F-F,F1F,
F)*&F=F,)\"\"$F0F,F,*&\"\"%F,)FBF0F,F," }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "This means" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "a(n)=s
oln0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,.*(,&%\"AG!\"
\"*&#\"\"\"\"\"#F/%\"BGF/F/F/,&F'F/F/F/F/)F0F'F/F/*&,&F1F.*&F0F/F+F/F,
F/F3F/F,*&,&F'#!#<F0#\"#<F0F,F/F3F/F/*&#\"\"*F0F/F3F/F,*&F?F/)\"\"$F'F
/F/*&\"\"%F/)FDF'F/F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 187 "Very convenient, but not very illuminati
ng! After all, of what importance is the solution to a contrived probl
em? It is the underlying ideas and the method of solution that are imp
ortant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
451 "If we were doing this example by hand we would first find the cha
racteristic polynomial of the associated homogeneous equation. Then we
 would observe the characteristic roots are  2, 2. In particular, 3  a
nd  4  are not characteristic. Thus we would expect a particular solut
ion of the form  A 3^n + B 4^n. We would substitute this trial solutio
n and find  A = 9 and  B = 4  (as we see above). This is not so bad, b
ut it can involve a lot of algebra. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 253 "In this worksheet we get Maple to do t
he algebra. Nothing we do here is needed if we just want a solution! T
he purpose of this worksheet is to provide support for hand calculatio
n so we can see how things work without being distracted by algebra er
rors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 
"It will be convenient, but not necessary, to have a procedure to norm
alize a polynomial, that is, to divide by the leading coefficient to m
ake the polynomial monic. The  degree()  function fails easily so  mon
ic()  is a bit tricky to get to work. In addition we have to simplify \+
the result to avoid problems with  roots()  later." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "monic:=proc
(p,z)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "  local k,c,q;" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 "  q:=sort(collect(p,z),z);" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 27 "  k:=degree(simplify(q),z);" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 18 "  c:=coeff(q,z,k);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 24 "  return(simplify(q/c));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "en
d:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
263 35 "Method of Undetermined Coefficients" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "The procedures present
ed here are not very robust. In particular we assume our linear recurr
ence equations are in normal form." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "`a(n) = c[1]*a(n-1) + ... +
c[m]*a(n-m) + f(n)`;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%Ma(n)~=~c[1]*
a(n-1)~+~...~+c[m]*a(n-m)~+~f(n)G" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "The labels  a,  n,  c,  m,  f  ca
n of course be \"anything\" but otherwise any other form will likely c
ause an error." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 82 "The method of undetermined coefficients which we will ill
ustrate below tells us if" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f(n) = P(n)*b^n;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/-%\"fG6#%\"nG*&-%\"PGF&\"\"\")%\"bGF'F+" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "where  P(
n)  is a polynomial of degree  t  then there is a particular solution \+
of the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 20 "a(n) = n^s*Q(n)*b^n;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"aG6#%\"nG*()F'%\"sG\"\"\"-%\"QGF&F+)%\"bGF'F+" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 168 "where  Q
(n)  is a polynomial of degree  t  with coefficients to be determined,
  s = 0  if  b  is not a characteristic root and  s  is the multiplici
ty of  b  otherwise." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 119 "If we combine this method with the superposition princ
iple we can solve quite a few simple linear recurrence equations." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 35 "T
he Associated Homogeneous Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 127 "First we need a procedure to compute the
 associated homogeneous recurrence equation, that is, to remove the in
homogeneous term." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 14 "hom:=proc(eqn)" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "  local inhom,a;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "  a:=op(0,lhs(eqn));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "  inhom
:=simplify(subs(a=0,rhs(eqn)));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "
  return(lhs(eqn)=simplify(rhs(eqn)-inhom));" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 4 "end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 15 "Let's check it." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "hom(eqn0);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,&-F%6#,&F'\"\"\"F,!\"\"\"\"%*&F.
F,-F%6#,&F'F,\"\"#F-F,F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 258 29 "The Characteristic Polynomial" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 174 "Next we want t
o compute the characteristic polynomial of a homogeneous recurrence re
lation. We normalize it to be monic for aesthetic reasons (and also so
 it will be unique)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 19 "cxpoly:=proc(eqn,z)" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "  local p,q,n,k,a;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "  n:=op(1,lhs(eqn));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "  a:=un
apply(lhs(eqn),n);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "  p:=subs(a=(
k->z^k),lhs(eqn)-rhs(eqn));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "  p:
=sort(simplify(p),z);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "  q:=op(p)
[nops(p)];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "  p:=simplify(p/q);" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "cpoly:=(eqn,z)->monic(cxpoly(eqn,z),z):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Let's che
ck it." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "cpoly(hom(eqn0),z); roots(%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(*$)%\"zG\"\"#\"\"\"F(*&\"\"%F(F&F(!\"\"F*F(" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#7#7$\"\"#F%" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 309 "The return from  roots()
  here means that  2  is  a root of multiplicity  2.  In particular  3
  and  4  are not a characteristic roots and so we have a paricular so
lution  eqn0  of the form  a(n) = A 3^n + B 4^n.  To compute  A  and  \+
B  we substitue our trial solution in  eqn0  and then solve for  A  an
d  B." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 30 "Substitution of Trial So
lution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "rsubs:=proc(sb,eqn)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 12 "  local e,n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "  n:=op(1,lhs
(sb));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "  e:=subs(unapply(lhs(sb)
,n)=unapply(rhs(sb),n), eqn);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "  \+
simplify(e);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Let's try
 it" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "subexpr:=rsubs(a(n)=A*3^n+B*4^n, eqn0); " }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%(subexprG/,&*&%\"AG\"\"\")\"\"$%\"nGF)F)*&%\"
BGF))\"\"%F,F)F),*F'#\"\")\"\"**(F+F)F.F))F0,&F,F)F)!\"\"F)F)F*F)F/F)
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
142 "The result of the substitution is an identity in  n.  Thus we tak
e special values of  n  to obtain a set of equations to solve for  A  \+
and  B." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "xeqn0:=\{subs(n=0,subexpr),subs(n=1,subexpr)\};" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&xeqn0G<$/,&%\"AG\"\"$*&\"\"%\"\"\"%
\"BGF,F,,(F(#\"\")F)*&F)F,F-F,F,\"\"(F,/,&F(F,F-F,,(F(#F0\"\"**&#F)F+F
,F-F,F,\"\"#F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(xeq
n0,\{A,B\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/%\"BG\"\"%/%\"AG\"
\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
260 9 "Example 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 41 "eqn1:=a(n)=4*a(n-1)-4*a(n-2)+n*2^n+n*5^n;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn1G/-%\"aG6#%\"nG,*-F'6#,&F)\"\"
\"F.!\"\"\"\"%*&F0F.-F'6#,&F)F.\"\"#F/F.F/*&F)F.)F5F)F.F.*&F)F.)\"\"&F
)F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eqn1h:=hom(eqn1);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&eqn1hG/-%\"aG6#%\"nG,&-F'6#,&F)
\"\"\"F.!\"\"\"\"%*&F0F.-F'6#,&F)F.\"\"#F/F.F/" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 34 "cp1:=cpoly(eqn1h,z); roots(cp1,z);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$cp1G,(*$)%\"zG\"\"#\"\"\"F**&\"\"%F*F(F*!
\"\"F,F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7$\"\"#F%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "Thus  2 \+
 is characteristic with multiplicity  2,  and  5  is noncharacteristic
. Hence we expect a particular solution of the form" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "trial1:=a
(n)=n^2*(A1+A2*n)*2^n + (A3+A4*n)*5^n;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%'trial1G/-%\"aG6#%\"nG,&*()F)\"\"#\"\"\",&%#A1GF.*&%#A2GF.F)F.
F.F.)F-F)F.F.*&,&%#A3GF.*&%#A4GF.F)F.F.F.)\"\"&F)F.F." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subex1:=rsubs(trial1,eqn1);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'subex1G/,**()%\"nG\"\"#\"\"\")F*F)F
+%#A1GF+F+*()F)\"\"$F+F,F+%#A2GF+F+*&)\"\"&F)F+%#A3GF+F+*(F3F+%#A4GF+F
)F+F+,6F'F+F.F+**\"\"'F+F,F+F1F+F)F+!\"\"*&)F*,&F)F+F+F+F+F-F+F;*(F:F+
F,F+F1F+F+*(#\"#;\"#DF+F3F+F5F+F+**FAF+F3F+F7F+F)F+F+*&#\"#7FCF+*&F3F+
F7F+F+F;*&F)F+F,F+F+*&F)F+F3F+F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "subeq1:=\{seq(subs(n=k,subex1),k=0..3)\};" }}{PARA 
12 "" 1 "" {XPPMATH 20 "6#>%'subeq1G<&/%#A3G,*%#A1G!\"#*&\"\"'\"\"\"%#
A2GF-F-*&#\"#;\"#DF-F'F-F-*&#\"#7F2F-%#A4GF-!\"\"/,*F)\"\"#*&F:F-F.F-F
-*&\"\"&F-F'F-F-*&F=F-F6F-F-,,F)F**&F:F-F.F-F-*&#F1F=F-F'F-F-*&#\"\"%F
=F-F6F-F-\"\"(F-/,*F)F1*&\"#KF-F.F-F-*&F2F-F'F-F-*&\"#]F-F6F-F-,,F)\"
\")*&FOF-F.F-F-*&F1F-F'F-F-*&\"#?F-F6F-F-\"#eF-/,*F)\"#s*&\"$;#F-F.F-F
-*&\"$D\"F-F'F-F-*&\"$v$F-F6F-F-,,F)\"#c*&\"$?\"F-F.F-F-*&\"#!)F-F'F-F
-*&\"$!=F-F6F-F-\"$*RF-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "
parm1:=solve(subeq1,\{A1,A2,A3,A4\});" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%&parm1G<&/%#A2G#\"\"\"\"\"'/%#A3G#!$+\"\"#F/%#A4G#\"#D\"\"*/%#A1G
#F)\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln1:=rhs(subs
(parm1,trial1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&soln1G,&*()%\"n
G\"\"#\"\"\",&#F*F)F**&#F*\"\"'F*F(F*F*F*)F)F(F*F**&,&#!$+\"\"#FF**&#
\"#D\"\"*F*F(F*F*F*)\"\"&F(F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 57 "Let's compare our solution with the \+
direct Maple solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "msoln1:=rsolve(eqn1,a);" }}{PARA 
12 "" 1 "" {XPPMATH 20 "6#>%'msoln1G,2*(,&-%\"aG6#\"\"!!\"\"*&#\"\"\"
\"\"#F/-F)6#F/F/F/F/,&%\"nGF/F/F/F/)F0F4F/F/*&,&F1F.*&F0F/F(F/F,F/F5F/
F,**F3F/,&F4F.F/F/F/,&F/F/*&#F/\"\"$F/F4F/F/F/F5F/F/*(F3F/F:F/F5F/F,*&
,&F4#!#V\"#=#\"#VFDF,F/F5F/F/*&#\"$H$\"#aF/F5F/F/*&,&F4#\"#D\"\"*FMF/F
/)\"\"&F4F/F/*&#\"$v\"\"#FF/FPF/F," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "diff1:=simplify(soln1-msoln1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&diff1G,,*()\"\"#%\"nG\"\"\"-%\"aG6#\"\"!F*F)F*F**&F'
F*F+F*!\"\"*()F(,&F)F*F*F0F*-F,6#F*F*F)F*F0*(#\"#P\"#=F*F)F*F'F*F**&#
\"$+\"\"#FF*F'F*F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 80 "It remains to show all the terms here are solutio
ns of the homogeneous equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "sub1:=rsubs(a(n)=diff1,eqn1h
): lhs(sub1)-rhs(sub1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 9 "Ex
ample 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 72 "eqn2:=a(n)=9*a(n-1)-9*a(n-2)-41*a(n-3)+42*a(n-4)+(2+5
*n^3)*7^n -2+n*4^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn2G/-%\"aG
6#%\"nG,0-F'6#,&F)\"\"\"F.!\"\"\"\"**&F0F.-F'6#,&F)F.\"\"#F/F.F/*&\"#T
F.-F'6#,&F)F.\"\"$F/F.F/*&\"#UF.-F'6#,&F)F.\"\"%F/F.F.*&,&F5F.*&\"\"&F
.)F)F;F.F.F.)\"\"(F)F.F.F5F/*&F)F.)FAF)F.F." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "eqn2h:=hom(eqn2);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%&eqn2hG/-%\"aG6#%\"nG,*-F'6#,&F)\"\"\"F.!\"\"\"\"**&F0F.-F'6#,&F)
F.\"\"#F/F.F/*&\"#TF.-F'6#,&F)F.\"\"$F/F.F/*&\"#UF.-F'6#,&F)F.\"\"%F/F
.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "cp2:=cpoly(eqn2h,z);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$cp2G,,*$)%\"zG\"\"%\"\"\"F**&\"
\"*F*)F(\"\"$F*!\"\"*&F,F*)F(\"\"#F*F**&\"#TF*F(F*F*\"#UF/" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "roots2:=roots(cp2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%'roots2G7&7$\"\"\"F'7$\"\"$F'7$\"\"(F'7$!\"#F'" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
141 "We note  7  and  1  are characteristic roots of multiplicity  1, \+
 and  4  is not a characteristic root. Thus we expect a solution of th
e form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 62 "trial2:=a(n)=n*(A1+A2*n+A3*n^2+A4*n^3)*7^n+n*A5+(A6+A
7*n)*4^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'trial2G/-%\"aG6#%\"nG,
(*(F)\"\"\",*%#A1GF,*&%#A2GF,F)F,F,*&%#A3GF,)F)\"\"#F,F,*&%#A4GF,)F)\"
\"$F,F,F,)\"\"(F)F,F,*&F)F,%#A5GF,F,*&,&%#A6GF,*&%#A7GF,F)F,F,F,)\"\"%
F)F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subex2:=rsubs(tri
al2,eqn2);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'subex2G/,0*(%\"nG\"\"
\")\"\"(F(F)%#A1GF)F)*()F(\"\"#F)F*F)%#A2GF)F)*()F(\"\"$F)F*F)%#A3GF)F
)*()F(\"\"%F)F*F)%#A4GF)F)*&F(F)%#A5GF)F)*&)F7F(F)%#A6GF)F)*(F<F)%#A7G
F)F(F)F),P!\"#F)*&\"#OF)F:F)!\"\"*&F/F)F*F)F)*(#\"\"*\"#kF)F<F)F?F)F)*
&#\"$;#\"$V$F)*&F*F)F,F)F)FD*&#\"#7\"#\\F)*&F*F)F0F)F)FD*(#\"$'yFMF)F*
F)F4F)F)*(\"#[F))F+,&F(F)F)FDF)F8F)FDF9F)*(#\"$b\"\"$G\"F)F<F)F=F)F)*&
F(F)F<F)F)*(\"\"&F)F*F)F2F)F)F'F)*&#\"$K%FMF)*(F*F)F0F)F(F)F)FD*&#\"$[
'FMF)*(F*F)F4F)F.F)F)FD*&#FCFRF)*(F*F)F4F)F(F)F)FD*&#\"$k)FMF)*(F*F)F8
F)F2F)F)FD*&#\"#sFRF)*(F*F)F8F)F.F)F)FD**#\"%WJFMF)F*F)F8F)F(F)F)**Ffn
F)F<F)F?F)F(F)F)F5F)F1F)F-F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 39 "subeq2:=\{seq(subs(n=k,subex2),k=0..6)\};" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%'subeq2G<)/,0%#A1G\"'%*eq*&\"(k`B%\"\"\"%#A2GF,F,*&\"
)%=7a#F,%#A3GF,F,*&\"*/JZ_\"F,%#A4GF,F,*&\"\"'F,%#A5GF,F,*&\"%'4%F,%#A
6GF,F,*&\"&wX#F,%#A7GF,F,,2\"*#z?t7F,*&\"#IF,F6F,!\"\"*&\"&O.$F,F<F,F,
*&\"'1=jF,F(F,F,*&\"('\\<LF,F-F,F,*&\")i;;<F,F0F,F,*&\")'H,z)F,F3F,F,*
&\"%g\\F,F9F,F,/,0F(\"&NS)*&\"'v,UF,F-F,F,*&\"(v35#F,F0F,F,*&\")vV]5F,
F3F,F,*&\"\"&F,F6F,F,*&\"%C5F,F9F,F,*&\"%?^F,F<F,F,,2\")2Ja5F,*&\"#JF,
F6F,FA*&\"%WjF,F<F,F,*&\"&^M(F,F(F,F,*&\"'>-JF,F-F,F,*&\"(\\QG\"F,F0F,
F,*&\"(2+D&F,F3F,F,*&\"%S7F,F9F,F,/,0F(\"%H5*&\"%(3$F,F-F,F,*&\"%h#*F,
F0F,F,*&\"&$yFF,F3F,F,*&\"\"$F,F6F,F,*&\"#kF,F9F,F,*&\"$#>F,F<F,F,,2\"
&\"=ZF,*&\"#LF,F6F,FA*&#\"$$[\"\"#F,F<F,F,*&\"$8)F,F(F,F,*&\"%2<F,F-F,
F,*&\"%fMF,F0F,F,*&\"%**pF,F3F,F,*&#\"$b\"F]qF,F9F,F,/,0F(\"%/'**&\"&;
%QF,F-F,F,*&\"'kO:F,F0F,F,*&\"'cYhF,F3F,F,*&\"\"%F,F6F,F,*&\"$c#F,F9F,
F,*&FZF,F<F,F,,2\"'WTxF,*&\"#KF,F6F,FA*&\"%w7F,F<F,F,*&\"%#4)F,F(F,F,*
&\"&Kd#F,F-F,F,*&\"&M&zF,F0F,F,*&\"'/FCF,F3F,F,*&\"$5$F,F9F,F,/,0F(\"#
)**&\"$'>F,F-F,F,*&\"$#RF,F0F,F,*&\"$%yF,F3F,F,*&F]qF,F6F,F,*&\"#;F,F9
F,F,*&FjrF,F<F,F,,2\"%)3#F,*&\"#MF,F6F,FA*&\"#TF,F<F,F,*&#\"$q%\"\"(F,
F(F,F,*&#\"$C%F]uF,F-F,F,*&\"#iF,F0F,F,*&#\"$'\\F]uF,F3F,F,*&#Fhq\"\")
F,F9F,F,/,0F(F]u*&F]uF,F-F,F,*&F]uF,F0F,F,*&F]uF,F3F,F,F6F,*&FcrF,F9F,
F,*&FcrF,F<F,F,,2\"#^F,*&\"#NF,F6F,FA*&#\"$t\"FjrF,F<F,F,*&#\"$F\"\"#
\\F,F(F,F,*&#FfvFjvF,F-F,FA*&#\"$H#FjvF,F0F,F,*&#\"$L#FjvF,F3F,FA*&#Fh
qFjrF,F9F,F,/F9,0F6!#O*&#\"\"*FcpF,F<F,F,*&#\"$;#\"$V$F,F(F,FA*&#\"#7F
jvF,F-F,FA*&#\"$'yF^xF,F0F,F,*&#\"#[F]uF,F3F,FA*&#Fhq\"$G\"F,F9F,F," }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "parm2:=solve(subeq2,\{A1,A
2,A3,A4,A5,A6,A7\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&parm2G<)/%#
A7G#!$G\"\"#F/%#A6G#\"$c#\"#\")/%#A4G#\"%:<\"$k)/%#A3G#!&0?\"\"%wx/%#A
2G#\"(D?V\"\"&7L*/%#A5G#!\"\"\"#=/%#A1G#!)*))G/&\"(;'z;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln2:=rhs(subs(parm2,trial2));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&soln2G,(*(%\"nG\"\"\",*#!)*))G/&\"(
;'z;F(*&#\"(D?V\"\"&7L*F(F'F(F(*&#\"&0?\"\"%wxF(*$)F'\"\"#F(F(!\"\"*&#
\"%:<\"$k)F()F'\"\"$F(F(F()\"\"(F'F(F(*&#F(\"#=F(F'F(F8*&,&#\"$c#\"#\"
)F(*&#\"$G\"\"#FF(F'F(F8F()\"\"%F'F(F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "Recall this is just a par
ticular solution. Here is Maple's complete solution for comparison" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "msoln2:=rsolve(eqn2,a);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'ms
oln2G,F-%\"aG6#\"\"\"#F)\"#O*&#\"\"(\"\"'F)-F'6#\"\"!F)F)*&#\"\"#\"\"*
F)-F'6#F5F)!\"\"*&F*F)-F'6#\"\"$F)F)*&,*F0#F.\"#?*&#F6\"#SF)F&F)F9*&#F
=FAF)F7F)F9*&#F)FDF)F;F)F)F))F=%\"nGF)F9*&,*F&#!#J\"$N\"*&#F.\"#XF)F0F
)F)*&#\"#6FOF)F7F)F)*&#F)FOF)F;F)F9F))!\"#FJF)F)*&,*F&#\"\"&\"$;#*&#F)
F+F)F0F)F9*&#F)\"$3\"F)F7F)F)*&#F)FhnF)F;F)F9F))F.FJF)F9*&#F)\"#=F)FJF
)F9#\")8Xl5\"%wxF9*&#\"(2Dn'\"%?^F)FIF)F)*&#\").#zW*\"'X_HF)FXF)F)*.#
\"%:<F+F),&FJF)F)F)F),&FJ#F)F5F)F)F),&F)F)*&#F)F=F)FJF)F)F),&FJ#F)\"\"
%F)F)F)F`oF)F)*&#\"'bj;\"%'H\"F)**FbpF)FcpF)FepF)F`oF)F)F9**#\"(&)o#p
\"&cm%F)FbpF)FcpF)F`oF)F)*&,&FJ#!**>>f>\"(;'z;#\"**>>f>FhqF9F)F`oF)F)*
&#\",(=e^`7\")whYgF)F`oF)F9*&,&FJ#!$G\"\"#F#\"$G\"FcrF9F))FjpFJF)F)*&#
\"$S'\"#\")F)FfrF)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT 262 14 "Bernoulli Sums" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The Bernoulli sum" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Sum
(k^m,k=1..n); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$SumG6$)%\"kG%\"mG
/F';\"\"\"%\"nG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 42 "is the solution of the recurrence equation" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 37 "eqn3:=a(n)=a(n-1)+n^m; init3:=a(0)=0;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%eqn3G/-%\"aG6#%\"nG,&-F'6#,&F)\"\"\"F.!\"\"F.)F)%\"m
GF." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&init3G/-%\"aG6#\"\"!F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Sin
ce " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "cp3:=cpoly(hom(eqn3),z);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$cp3G,&%\"zG\"\"\"F'!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 "we see  1  is characteri
stic with multiplicity  1.  Thus we expect a particular solution of th
e form  a(n) = n*P(n)  where  P(n)  is a polynomial of degree  m.  The
 solution of the homogeneous equation is obviously just a constant. Th
us the general solution is  c + n*P(n).  Substituting the initial cond
ition yields  c = 0. Thus" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Sum(k^m,k=1..n)=n*P(n); " }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$)%\"kG%\"mG/F(;\"\"\"%\"nG*&F-F,-%
\"PG6#F-F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 " where  P(n)  is a
 polynomial of degree  m. As above we determine the coefficients of  P
(n)  from a " }{TEXT 276 6 "finite" }{TEXT -1 18 " number of cases. " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "For  m \+
= 4  we have" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 46 "trial3:=a(n)=n*(A1+A2*n+A3*n^2+A4*n^3+A5*n^4);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'trial3G/-%\"aG6#%\"nG*&F)\"\"\",,
%#A1GF+*&%#A2GF+F)F+F+*&%#A3GF+)F)\"\"#F+F+*&%#A4GF+)F)\"\"$F+F+*&%#A5
GF+)F)\"\"%F+F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "subex3
:=rsubs(trial3,subs(m=4,eqn3));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'
subex3G/*&%\"nG\"\"\",,%#A1GF(*&%#A2GF(F'F(F(*&%#A3GF()F'\"\"#F(F(*&%#
A4GF()F'\"\"$F(F(*&%#A5GF()F'\"\"%F(F(F(,LF,F(F.!\"\"F6F:*(\"\"&F(F6F(
F7F(F:*(F0F(F,F(F'F(F:*(F8F(F2F(F'F(F:F2F(*(F4F(F.F(F'F(F(*(\"\"'F(F2F
(F/F(F(*(\"#5F(F6F(F3F(F(*(FCF(F6F(F/F(F:*&F'F(F*F(F(*&F,F(F/F(F(*&F.F
(F3F(F(*&F2F(F7F(F(*&F6F()F'F<F(F(*(F4F(F.F(F/F(F:*(F8F(F2F(F3F(F:*(F<
F(F'F(F6F(F(F*F:*$F7F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 
"subeq3:=\{seq(subs(n=k,subex3),k=0..4)\};" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%'subeq3G<'/,,%#A1G\"\"$*&\"\"*\"\"\"%#A2GF,F,*&\"#FF,
%#A3GF,F,*&\"#\")F,%#A4GF,F,*&\"$V#F,%#A5GF,F,,.F-\"\"%*&\"\")F,F0F,F,
*&\"#KF,F6F,F,*&\"#;F,F3F,F,*&\"\"#F,F(F,F,F2F,/,,F(F@*&F8F,F-F,F,*&F:
F,F0F,F,*&F>F,F3F,F,*&F<F,F6F,F,,.F-F,F0F,F6F,F3F,F(F,F>F,/\"\"!,,F-F,
F0!\"\"F6FKF3F,F(FK/,,F(F,F-F,F0F,F3F,F6F,F,/,,F(F8*&F>F,F-F,F,*&\"#kF
,F0F,F,*&\"$c#F,F3F,F,*&\"%C5F,F6F,F,,.F-F+*&F/F,F0F,F,*&F5F,F6F,F,*&F
2F,F3F,F,*&F)F,F(F,F,FTF," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
38 "parm3:=solve(subeq3,\{A1,A2,A3,A4,A5\});" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&parm3G<'/%#A2G\"\"!/%#A5G#\"\"\"\"\"&/%#A3G#F,\"\"$/
%#A1G#!\"\"\"#I/%#A4G#F,\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 31 "soln3:=rhs(subs(parm3,trial3));" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%&soln3G*&%\"nG\"\"\",*#!\"\"\"#IF'*&#F'\"\"$F')F&\"\"#F'F'*&#F'F0
F')F&F.F'F'*&#F'\"\"&F')F&\"\"%F'F'F'" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Of course, we can evaluat
e the Bernoulli sums directly in Maple" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "bern4:=sum(k^4,k=1..n);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&bern4G,,*$),&%\"nG\"\"\"F*F*\"
\"&F*#F*F+*&#F*\"\"#F**$)F(\"\"%F*F*!\"\"*&#F*\"\"$F*)F(F6F*F**&#F*\"#
IF*F)F*F3#F*F:F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplif
y(soln3-bern4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 8 "Problems
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Use t
he method of undetermined coefficients as above to find particular sol
utions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 9 "Problem 1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "a(
n)=4*a(n-1)+4*a(n-2)+1+n+2^n+n^2*3^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#/-%\"aG6#%\"nG,.-F%6#,&F'\"\"\"F,!\"\"\"\"%*&F.F,-F%6#,&F'F,\"\"#F-
F,F,F,F,F'F,)F3F'F,*&)F'F3F,)\"\"$F'F,F," }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 9 "Problem 2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "a(
n)=8*a(n-1)-17*a(n-2)-6*a(n-3)+44*a(n-4)-8*a(n-5)-32*a(n-6)+n*2^n+n*4^
n+n*(-1)^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,4-F%6#,&
F'\"\"\"F,!\"\"\"\")*&\"#<F,-F%6#,&F'F,\"\"#F-F,F-*&\"\"'F,-F%6#,&F'F,
\"\"$F-F,F-*&\"#WF,-F%6#,&F'F,\"\"%F-F,F,*&F.F,-F%6#,&F'F,\"\"&F-F,F-*
&\"#KF,-F%6#,&F'F,F6F-F,F-*&F'F,)F4F'F,F,*&F'F,)F@F'F,F,*&F'F,)F-F'F,F
," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Problem 3" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 14 "a(n)=a(n-4)+n;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"aG6#%\"nG,&-F%6#,&F'\"\"\"\"\"%!\"\"F,F'F," }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Problem 4" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "a(n)=a(n-2)+n^4*2^n;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"aG6#%\"nG,&-F%6#,&F'\"\"\"\"\"#!\"\"F,*&)F'\"\"%F,
)F-F'F,F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Problem 5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a(n)=-a(n-2)+n^4*2^n;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,&-F%6#,&F'\"\"\"\"\"#!\"\"F.*&)F'
\"\"%F,)F-F'F,F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Problem 6" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a(n)=a(n-2)+(n^4-n^2)*2^n;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,&-F%6#,&F'\"\"\"\"\"#
!\"\"F,*&,&*$)F'\"\"%F,F,*$)F'F-F,F.F,)F-F'F,F," }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 9 "Problem 7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
50 "a(n) = 9*a(n-1)-27*a(n-2)+27*a(n-3)+n^3*(3^n+4^n);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#/-%\"aG6#%\"nG,*-F%6#,&F'\"\"\"F,!\"\"\"\"**&\"#FF
,-F%6#,&F'F,\"\"#F-F,F-*&F0F,-F%6#,&F'F,\"\"$F-F,F,*&)F'F9F,,&)F9F'F,)
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