Question

Help me please with this problem!!

Answer 1

Let p(n) be the statement \(n!\ge 2^{n-1}\)
Consider p(1),
LS = 1! =1
RS = \(2^{1-1} = 2^0 = 1 = LS\)
So, p(1) is true.
Suppose p(k) is true for some positive integers k.
LS of p(k+1) = (k+1)!
= (k+1)(k!)
\(\ge (k+1)(2^{k-1})\) (since we suppose \(k! \ge 2^{k-1}\) is true)
Can you continue from here?

Answer 2

If you dont mind Castillo and you please guide me through this whole problem because these are really tough problems for me :(

Answer 3

Since k is a positive integer, k+1 must be greater than or equal to 2, what conclusion can you draw from here?

Answer 4

so are we taking (k+1)(k+1) is greater than or equal to 2?

Answer 5

Hmmm.. Think about it.
If k=1, k+1 = 1+ 1 = 2
So, \((k+1)(2^{k-1}) = 2(2^{k-1}) = 2^k\)

If k >1, k+1 > 2
So, what is the relation between \((k+1)(2^{k-1})\) and \(2^k\)?

Answer 6

omg I m just too confuse :( /i dont know what I ll do in exam

Answer 7

Which part are you confused at?

Answer 8

isnt it equivalence relation?

Answer 9

They are not equal...
Options you can have: >, <, =.

Answer 10

ok so they are greater than 2^k

Answer 11

Yes.
So, combine the two cases together. If k is a positive integer, \((k+1)(2^{k-1})\ge2^k\)
So far so good?

Answer 12

yup

Answer 13

So, you can write the next step and it's done!

Answer 14

so If I m writing next step then I ll include this (k+1)(2^k-1)greater than 2^k right?

Answer 15

Yes.

Answer 16

ok thank you :) for your help.

Answer 17

Welcome :)