1. 5 (a)
Tutorial 2 solutions
January 2021
(b) Wehave􏰆nk=0akbn−k =3nifn≥1,and1forn=0,andso A(x)B(X)=1+􏰇3nxn =1+3x􏰇(n+1)xn =1+
n≥1 n≥0
3x . (1 − x)2
(c) Wehave􏰆n akbn−k =⌈n⌉whennisoddand0whenniseven. k=0 2
From here finding the closed form is a bit tricky. Probably the easiest way to go about doing this is to notice that 􏰆 (n + 1)x2n = ( 1 )2, but proving this
n≥0 1−xr
isn’t really any different than just solving the problem as in 6 part (c)…
2. 6 (a)
(b) WeseethatA(x)􏰆
n≥0
(n+1)xn = 1 andB(x)=1+x+x2.Consequently (1−x)2
1+x+x2 A(X)B(X)= (1−x)2 .
(c)NotethatA(x)=􏰆 x2n = 1 andB(x)=􏰆 x2n+1 =x􏰆 x2n =
x . Consequently, 1−x2
n≥0
1−x2 n≥0 n≥0 A(x)B(x) = x
(1 − x2)2
1

1. 5 (a)
Tutorial 2 solutions
January 2021
(b) Wehave􏰆nk=0akbn−k =3nifn≥1,and1forn=0,andso A(x)B(X)=1+􏰇3nxn =1+3x􏰇(n+1)xn =1+
n≥1 n≥0
3x . (1 − x)2
(c) Wehave􏰆n akbn−k =⌈n⌉whennisoddand0whenniseven. k=0 2
From here finding the closed form is a bit tricky. Probably the easiest way to go about doing this is to notice that 􏰆 (n + 1)x2n = ( 1 )2, but proving this
n≥0 1−xr
isn’t really any different than just solving the problem as in 6 part (c)…
2. 6 (a)
(b) WeseethatA(x)􏰆
n≥0
(n+1)xn = 1 andB(x)=1+x+x2.Consequently (1−x)2
1+x+x2 A(X)B(X)= (1−x)2 .
(c)NotethatA(x)=􏰆 x2n = 1 andB(x)=􏰆 x2n+1 =x􏰆 x2n =
x . Consequently, 1−x2
n≥0
1−x2 n≥0 n≥0 A(x)B(x) = x
(1 − x2)2
1