Answer
a. f(1) = 3, f(2) = 5, f(3) = 7, f(4) = 9
b. f(1) = 3, f(2) = 9, f(3) = 27, f(4) = 81
c. f(1) = 2, f(2) = 4, f(3) = 16, f(4) = 65536
d. f(1) = 3, f(2) = 13, f(3) = 183, f(4) = 33673
Work Step by Step
a. f(n+1) = f(n) + 2, f(0) = 1
f(0+1) --> f(1) = f(0) + 2 = 1 + 2 = 3
--> f(1) = 3
f(1+1) --> f(2) = f(1) + 2 = 3 + 2 = 5
--> f(2) = 5
f(2+1) --> f(3) = f(2) + 2 = 5 + 2 = 7
--> f(3) = 7
f(3+1) --> f(4) = f(3) + 2 = 7 + 2 = 9
--> f(4) = 9
b. f(n+1) = 3×f(n), f(0) = 1
f(0+1) --> f(1) = 3×f(0) = 3×1 = 3
--> f(1) = 3
f(1+1) --> f(2) = 3×f(1) = 3×3 = 9
--> f(2) = 9
f(2+1) --> f(3) = 3×f(2) = 3×9 = 27
--> f(3) = 27
f(3+1) --> f(4) = 3×f(3) = 3×27 = 81
--> f(4) = 81
c. f(n+1) = $2^{f(n)}$, f(0) = 1
f(0+1) --> f(1) = $2^{f(0)}$ = $2^{1}$ = 2
--> f(1) = 2
f(1+1) --> f(2) = $2^{f(1)}$ = $2^{2}$ = 4
--> f(2) = 4
f(2+1) --> f(3) = $2^{f(2)}$ = $2^{4}$ = 16
--> f(3) = 16
f(3+1) --> f(4) = $2^{f(3)}$ = $2^{16}$ = 65536
--> f(4) = 65536
d. f(n+1) = $f(n)^{2}$ + f(n) + 1, f(0) = 1
f(0+1) --> f(1) = $f(0)^{2}$ + f(0) + 1 = $1^{2}$ + 1 + 1 = 3
--> f(1) = 3
f(1+1) --> f(2) = $f(1)^{2}$ + f(1) + 1 = $3^{2}$ + 3 + 1 = 13
--> f(2) = 13
f(2+1) --> f(3) = $f(2)^{2}$ + f(2) + 1 = $13^{2}$ + 13 + 1 = 183
--> f(3) = 183
f(3+1) --> f(4) = $f(3)^{2}$ + f(3) + 1 = $183^{2}$ + 183 + 1 = 33673
--> f(4) = 33673
