Solution to Problem 1
(a) The characteristic equation is r2 − 1 = 0.
Thus, r = 1, −1 and the solution can be written as an = α1(1)n + α2(−1)n.
Fromtheinitialconditions,a0 =5=α1+α2 anda1 =−1=α1−α2.
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Wegetα1 =2andα2 =3.
Finally, an = 2 + 3 · (−1)n.
(b) It is a degree two linear homogeneous recurrence relation. Its characteristic equation is r2 + 4r − 5 = 0.
Wegetr=1,−5andan =α1(1)n+α2(−5)n. Fromtheinitialconditions,a0 =2=α1+α2 anda1 =8=α1−5α2. Bysolving,α1 =3andα2 =−1.
The solution is an = 3 − (−5)n.
Solution to Problem 2
This is a linear homogeneous recurrence relation with degree 3. The characteristic equation is r3 − 2r2 − r + 2 = 0.
The characteristic roots are obtained as r = 1,−1,2. These are distinct roots. Thus, the solution can be written as α1(1)n + α2(−1)n + α3(2)n.
Fromtheinitialconditions,a0 =3=α1+α2+α3,a1 =6=α1−α2+2α3,and a2 = 0 = α1 + α2 + 4α3.
We get α1 = 6, α2 = −2, and α3 = −1. The solution is 6 − 2(−1)n − (2)n. 1/3
Solution to Problem 3
The characteristic equation is r3 − 6r2 + 12r − 8 = 0. It can be written as r3 −3·r2 ·2+3·r·22 −23 =0.
Thus, we have, (r − 2)3 = 0 or r = 2 with multiplicity 3. In this case, the general form of the solution is (α1 + nα2 + n2α3)2n.
From the initial conditions, a0 = −5 = α1, a1 = 4 = 2α1 +2α2 +2α3, and a2 =88=4α1 +8α2 +16α3.
Weobtainα1 =−5,α2 =1/2,andα3 =13/2.
Finally, the solution is (−5 + n/2 + 13n2/2)2n = −5 · 2n + n · 2n−1 + 13n2 · 2n−1.
Solution to Problem 4
(a) Taking an = n2n, we get, 2an−1 +2n = 2(n−1)2n−1 +2n = n2n = an. This completes the proof.
(b) The associated homogeneous recurrence relation is with degree 1. Its root is r = 2. Therefore, the general form of the solution is, an = α2n + n2n.
(c) From the given initial condition, a0 = 2 = α or α = 2. The solution is (n + 2)2n
Solution to Problem 5
At first, we need to know the roots of the associated homogeneous recurrence rela- tion.
The characteristic equation is r3 − 6r2 + 12r − 8 = 0. Its root is r = 2 with multi- plicity 3. Now the solutions can be obtained using Theorem 6.
(a) For F (n) = 3 · (1)n, we observe that 1 is not a root of the associated homoge- neous equation. Therefore, the particular solution will be in the form p0.
(b) For F (n) = n2 · (1)n, we observe that 1 is not a root of the associated homoge- neous equation. Therefore, the particular solution will be in the form p2n2+p1n+p0.
(c) For F(n) = 2n, we observe that 2 is a root of the associated homogeneous equation with multiplicity 3. Therefore, the particular solution will be in the form n3p02n.
(d) For F(n) = n22n, we observe that 2 is a root of the associated homogeneous equation with multiplicity 3. Therefore, the particular solution will be in the form n3(p2n2 + p1n + p0)2n.
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