Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342.

Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Chapter1: Introduction
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Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342.

342
8 Sequences and Series
is given in two parts:
if a = 1,
for Vn E N;
for Vn e N.
Sn = I+nc
%3|
if a + 1,
Sn = a"A+
1
- a
When a = 1, any particular solution is obtained by determining a specific,
numerical value for I. In fact, a particular solution is determined by a specific,
numerical value J for any (particular) entry, S;. Solving the equation
J = I+jc for I,
I = J – jc.
// since S; = I+ jc
// where So =I
we get
// One particular “particular solution" has I = 0.
When a + 1, any particular solution is obtained by determining a specific,
numerical value for A; if the starting value I is given, then A = I -
In fact,
1 - a
a particular solution is determined by a specific, numerical value J for any
(particular) entry, S;. Solving the equation
J = Aa' +
for A,
- a
we get
A =
aJ
// But what if a = 0?
1 al
// One particular “particular solution" has A = 0.
Example 8.2.1: The Towers of Hanoi
The recurrence equation for the number of moves in the Towers of Hanoi
problem is a first-order linear recurrence equation:
T = 2T,-1+1.
1
2 and c =
= -1, and any sequence T that satisfies
2
Here a =
1, so
1
1
a
this RE is given by the formula
T, = 2"[1 – (–1)] +(-1)
= 2"[I + 1] – 1.
Assuming T has domain N and denoting To by I, we saw at the beginning of this
chapter several particular solutions:
= (0, 1,3, 7, 15, 31,...);
T = (2,5,11, 23, 47, 95, ...);
// T, = 2" [0+1] – 1 = 2"
if I = 0, then
if I = 2, then
if I = 4, then
if I = -1, then T = (-1,–1,-1,-1,–1,...). // T, = 2"(–1+1] – 1=
- 1.
// Tn = 2"[2+1] – 1 = 3 × 2" – 1.
T = (4,9, 19, 39, 79, 159, ...); // T, = 2" [4 + 1] – 1 = 5 x 2" – 1.
-1.
Transcribed Image Text:342 8 Sequences and Series is given in two parts: if a = 1, for Vn E N; for Vn e N. Sn = I+nc %3| if a + 1, Sn = a"A+ 1 - a When a = 1, any particular solution is obtained by determining a specific, numerical value for I. In fact, a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = I+jc for I, I = J – jc. // since S; = I+ jc // where So =I we get // One particular “particular solution" has I = 0. When a + 1, any particular solution is obtained by determining a specific, numerical value for A; if the starting value I is given, then A = I - In fact, 1 - a a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = Aa' + for A, - a we get A = aJ // But what if a = 0? 1 al // One particular “particular solution" has A = 0. Example 8.2.1: The Towers of Hanoi The recurrence equation for the number of moves in the Towers of Hanoi problem is a first-order linear recurrence equation: T = 2T,-1+1. 1 2 and c = = -1, and any sequence T that satisfies 2 Here a = 1, so 1 1 a this RE is given by the formula T, = 2"[1 – (–1)] +(-1) = 2"[I + 1] – 1. Assuming T has domain N and denoting To by I, we saw at the beginning of this chapter several particular solutions: = (0, 1,3, 7, 15, 31,...); T = (2,5,11, 23, 47, 95, ...); // T, = 2" [0+1] – 1 = 2" if I = 0, then if I = 2, then if I = 4, then if I = -1, then T = (-1,–1,-1,-1,–1,...). // T, = 2"(–1+1] – 1= - 1. // Tn = 2"[2+1] – 1 = 3 × 2" – 1. T = (4,9, 19, 39, 79, 159, ...); // T, = 2" [4 + 1] – 1 = 5 x 2" – 1. -1.
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