Question

Prove by induction:

a) n! >= n*2^(n-2)
b) n! <= n^(n-1)

Thanx!

Answer 1

okay
n! is meant to n(n-1) so we got
n(n-1)>= n*2^(n-2)
by divide each sides by (n-1)
n>=2^(n-2) that's mean is correct subtitde by 2 for example
2>=2^0
2>=1

Answer 2

but i have to prove it by induction, so (n+1)! >=...

Answer 3

assume it to be true upto n
(n+1)! = n! (n+1) >= (n+1)*2^(n-1)
=> n! >=2^(n-1) => n * 2^(n-2) for n >= 2
therefore (n+1)! >= (n+1)*2^(n+1-2)

Answer 4

assume b) upto n ...
(n+1)! = n! (n+1) <= (n+1)^n = (n+1) (n+1)^(n-1) => n! <= (n+1)^(n-1)
which is true ...
therefoe
(n)! <= n^(n-1)

Answer 5

hash the second step how is it thata

Answer 6

sorry i didn't get u

Answer 7

n!(n+1) what is equal

Answer 8

where??

Answer 9

(n+1)! = (n+1) x n!

Answer 10

okay and n! can it simplify or not

Answer 11

b)
Base Case: n = 1;

1! = 1
n^(n-1) = 1 ^ 0 = 1
1 <= 1 is true.
Base Case holds//

Assume n = k holds;

hence:

k! <= k^(k-1) is true!

prove that n = k + 1 holds too.

LHS: Now (k+1)! = (k+1)k!
RHS: (k+1) ^ k

need to prove that:

(k+1)k! <= (k+1)^k

Divide by k+1 (since its positive):

k! <= (k+1)^(k-1)

Now since we know that k! <= k^(k-1) is true,
the right hand side can only grow with a k+1,
hence k! remains <= (k+1)^(k-1)

Answer 12

aaaah okok, i see now, thnx everybody!

Answer 13

thanx