(Prove by induction) Show that n!>2^n for n>=4 ?

Answer 1

n!>2^n for n>=4

i)Base case: n = 4
n! = 4*3*2*1 = 24
2^n= 2 ^4 = 16

n!>2^n is true

ii) Step by Induction

Consider for a n that the statement n!>2^n is true
Show for n+1

n+1! = (n+1) *n!
2^(n+1) = 2 *2^n

n+1! > 2^(n+1) ?
(n+1) *n! > 2 *2^n ?

We know that n! > 2^n and as n>=4, n+1 > 2 so:

(n+1) *n! > 2 *2^n is true

(n+1) ! > 2^(n+1) is true

And now we can conclude that n!>2^n is true for n>=4 by induction



Well I wish i help u

Kisses from Brazil.

Answer 2

(n+1)! = n! (n+1) > 2^n (n+1) > 2^n * 2 = 2^(n+1).

Answer 3

4! = 24
2^4 = 16

Therefore n! > 2^n for n = 4

(n+1)! = (n + 1) * n!
2^(n+1) = 2^n * 2

Since n! > 2^n and (n + 1) > 2 for all n >= 4,

(n+1) * n! > 2 * 2^n

Therefore, (n+1)! > 2^(n+1).

This proves that if it holds for n = 4, then it's also true for n = 5. And if it's true for n = 5, it's also true for n = 6, and so on...

Answer 4

See my answer to a question from KR four days ago. It started "if n (belongs to sign) N . . .