Question

Help with induction problem:

Answer 1

so you are just having a problem on the inductive step?

Answer 2

you need to show it for the first case n=1
now assume it is true for some integer k
actually write down what you are assuming
you might want to also write down what you want to show
look what you can do to the inductive step to get what you want to show

Answer 3

hint: recall an inequality still holds if you multiply both sides of it by a positive number

Answer 4

So multiply both sides by 2^(n+1)?

Answer 5

start with the assumption that it is true for some n.

Answer 6

n=1 is true.

Answer 7

okay now write down what you are assuming is true

Answer 8

So assuming that for all \[n \ge 1\]
It'll be true?

Answer 9

multiply by 2, an and notice k-2< k-1

Answer 10

when k is natural...

Answer 11

When k is natural, it'll always be true.

Answer 12

I'm saying write down what you are assuming is true:

\[\text{ Assume } 2^k \le 2^{k+1}-2^{k-1}-1 \text{ for some integer } k \ge 1 \]

Answer 13

do some manipulation to show that the following is true:
I'm saying take the assumption and show the following is true:

\[2^{k+1} \le 2^{(k+1)+1}-2^{(k-1)+1}-1 \]
let me rewrite this
you need to show:
\[2^{k+1} \le 2^{k+2}-2^k-1\]
you get to use the following assumption
\[\text{ Assume } 2^k \le 2^{k+1}-2^{k-1}-1 \text{ for some integer } k \ge 1\]
and any other math things you know like -2<-1 or if you have x>y then 2x>2y.

Answer 14

what can you do to the assumption to get to what you want!

Answer 15

i wonder how one can get from 2^k to 2^(k+1)

Answer 16

i wonder what 2^(k+1)/2^k is

Answer 17

i wonder if i can use law of exponents there to simplify that above thingy

Answer 18

thingy quotient whatever

Answer 19

I gotcha. I'll keep working on it now, and post if I run into trouble later.

Answer 20

and remember if x<y then 2x<2y
or for that mather mx<my for any m>0

Answer 21

\(2^n\le2^{n+1}-2^{n-1}-1\) is the assumption

what you want

\(2^{n+1}\le 2^{n+2}-2^n-1\)

so take \(2^n\le2^{n+1}-2^{n-1}-1\) and multiply it by \(2\)

\(2*2^n\le2*2^{n+1}-2*2^{n-1}-2*1\le ?\)

Answer 22

matter * not mather

Answer 23

So set k = k+1, and the assumption turns into what we want, right?

Answer 24

n = n+1, right? *

Answer 25

?

Answer 26

i'm not sure what you are saying

Answer 27

n=n+1????

Answer 28

2*2^n = 2^{n+1}

Answer 29

do that a bunch

Answer 30

Ohh, that makes more sense. Disregard what I said.