We saw in lecture a new kind of unbounded looping: recursion. In this problem, youll practice

translating a recurrence to code.

s(0) = 0

s(n) =s(n1) +n

s1(n) =n(n+ 1)2

p(0) = 10000

p(n) =p(n1) + 0.02p(n1)

p1(n) = (1.02)^n10000

b(1)= 2

b(2)=3

b(n) =b(n1) +b(n2)n= 3,4,5, . . .

c(1) = 9

c(n) = 9c(n1) + 10^n1c(n1)

d(0) = 1

d(n) = 3d(n1) + 1

d1(n) =3n+11/2