If P(n) :

Solution:

Given P(n) : 2n < n! ; n ∈ N,

for n = 1,

P(1) : 2' < 1!

i.e 2 < 1

Which is not true

for n = 2,

P(2) : 22 = 4 < 2!

i.e 4 < 2

Which is not true

for n = 3,

P(3) : 23 < 3!

i.e $8<1 \times 2 \times 3$

i.e $8<6$

Which is again not true.

for n = 4,

P(4) i.e 24 < 4!

i.e 16 < 24

i.e a true statement.

P(5) : 25 < 5!

i.e 32 < 1 × 2 × 3 × 4 × 5

i.e 32 < 120

Which is also true

$\therefore F(n): 2^{n}3$

$F(n): 2^{n}3$