What is the smallest positive integer for which the inequali

What is the smallest positive integer for which the inequality n!

Solution

1.
for n = 1,
LHS = n! = 1
RHS = ((n + 1) / 2)^n = ((1 + 1) / 2)^1 = 1

for n = 2,
LHS = n! = 2
RHS = ((n + 1) / 2)^n = ((2 + 1) / 2)^2 = (1.5)^2 = 2.25

So, LHS < RHS.
So, answer:
b. 2

2.
c. Basis step

3.
b. 5 + k*5, where k wil be decreased by 1 after every recursive call.

4.
e. It calls itself