Answer:
f(n) = 4 + 3(n - 1) for n≥1
f(n)= 3n+ 1 for n ≥ 1
f(1) = 4 ; f(n) = 3 +f(n -1) for n ≥2
Explanation:
f(n) = 4 + 3(n - 1) for n≥1
For n=1 ; f(1) = 4 + 3 * (1 - 1)
= 4 + 3*0
= 4 squares
For n = 2 ; f(2) = 4 + 3*1
= 4 + 3
= 7 squares
For n= 3 ; f(3) = 4 + 3*2
= 4 + 6
= 10 squares
For n = 4 ; f(4) = 4 + 3 * 3
= 4 + 9
= 13 squares
f(n)= 3n+ 1 for n ≥ 1
f(1) = 3 * 1 + 1 = 3 + 1 = 4 squares
f(2) = 3*2 + 1 = 6 + 1 = 7 squares
f(3) = 3*3 + 1 = 9 + 1 = 10 squares
f(4) = 3*4 + 1 =12 + 1 = 13 squares
f(1) = 4 ; f(n) = 3 +f(n -1) for n ≥2
f(2) = 3 + f(2-1)
= 3 + f(1)
= 3 + 4
= 7 squares
f(3) = 3 + f(2)
= 3 + 7
= 10 squares
f(4) = 3 + f(3)
= 3 + 10
= 13 squares