Question

Using math induction prove that (2n^3) less than or = (4^n). If N greater or equal 3

Answer 1

So you want to prove \[2n^3 \le 4n \text{ for } n \ge 3\]

Answer 2

show the base case first

Answer 3

that is the base case being n=3

Answer 4

it doesn't look like it holds for n=3

Answer 5

so you are done
because it is false

Answer 6

any questions?

Answer 7

or is that not what you meant?

Answer 8

i replaced n by 3 for 2n^3 gave me 54 and 4^3 gave me 81 soo in this case wouldn't it be true that the 4^n is greater for the case

Answer 9

So you are doing a different problem now?

Answer 10

I see 4n

Answer 11

oh my mistake i meant for 4^n, sorry

Answer 12

\[2n^3 \le 4^n \text{ for } n \ge 3\]

Answer 13

so if this was the inequality
should it is true for n=3
and yes this one would be true for the base case

Answer 14

exactly i got stuck when replacing the n for (k+1)

Answer 15

now you suppose
\[2k^3 \le 4^k \text{ for some integer k } \\ \text{ and we want to show } 2(k+1)^3 \le 4^{k+1}\]

Answer 16

hmm... now we think on different strategies

Answer 17

i got to that part and then i'm stuck what to do next

Answer 18

just thinking
\[2(k+1)^3 =2(k^3+3k^2+3k+1)\\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1)\]

Answer 19

\[2(k+1)^3 =2(k^3+3k^2+3k+1)\\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1) \\ \]
well we know k>=3

Answer 20

thought i had an idea
but it escaped me

Answer 21

just so we can think faster I will call @ganeshie8 here

Answer 22

sounds good

Answer 23

\[2(k+1)^3 =2(k^3+3k^2+3k+1)\\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1) \\ \\ \le 4^k+2(kk^2+kk+1) =4^k+2(k^3+k^2+1)\]

Answer 24

it looks like we can use our induction hypothesis again

Answer 25

\[2(k+1)^3 =2(k^3+3k^2+3k+1)\\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1) \\ \\ \le 4^k+2(kk^2+kk+1) =4^k+2(k^3+k^2+1) \\ \le 4^k+4^k+2(k^2+1)=2 \cdot 4^k+2(k^2+1)\]

Answer 26

we want to show 2(k^2+1) is less than or equal to 2*4^k if we can

Answer 27

because 2*4^k+2*4^k=4*4^k=4^(k+1)

Answer 28

so we can simplify this is a little show now that k^2+1<=4^k somehow

Answer 29

ok I see what you've done

Answer 30

i didn't finish it
there is still a little something for you to do

Answer 31

\[\forall k \ge 1, ~~k^2+1 \le k^3 +k^3 = 2k^3 \le 4^k\]

Answer 32

yep

Answer 33

ah the last piece is by induction hypothesis so its not k>=1. it should be k>=3

Answer 34

I wonder if there is a shorter faster way

Answer 35

I guess I could have left that one thing as 2(3k+1) instead of wrote <=2(kk+1)=2(k^2+1)

Answer 36

I kinda like it as 3k<kk<k^3 though

Answer 37

where would as 2(3k+1)? right before you factor 2(k+1)^3

Answer 38

\[2(k+1)^3 =2(k^3+3k^2+3k+1)\\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1) \\ \\ \le 4^k+2(kk^2+kk+1) =4^k+2(k^3+k^2+1) \\ \le 4^k+4^k+2(k^2+1)=2 \cdot 4^k+2(k^2+1) \]
second line at the end
the 2(3k^2+3k+1)
or 2(3k^2)+2(3k+1)

Answer 39

your pieces are a bit scattered
do you think you can piece it all together?
does anything not make sense

Answer 40

it's starting to make a lot more sense I just have to view all things that have been written so i don't get mixed up

Answer 41

i didn't know where i was going to go at first
sometimes you just have to play with ideas
i kinda think sometimes it helps to be a scatterbrain
I could be wrong about it

Answer 42

it being the scatterbrain part

Answer 43

i usually call @ganeshie8 when I get nervous and scared :p

Answer 44

no it's fine thank you very much , I needed help on this problem. the section where you wrote 4^k+2(3k^2...)you factored out the 3 and where did it go?

Answer 45

I will write a reason to the right or directly underneath each step

Answer 46

one sec

Answer 47

\[2(k+1)^3 =2(k^3+3k^2+3k+1) \text{ multiplied } \\ =2k^3+2(3k^2+3k+1) \le 4^k+2(3k^2+3k+1) \text{ use induction hypothesis } \\ \le 4^k+2(kk^2+kk+1) \text{ used } k \ge 3 \\ =4^k+2(k^3+k^2+1) \text{ did a little multipling } \\ \le 4^k+4^k+2(k^2+1) \text{ used induction hypothesis again } \\ =2\cdot 4^k+2(k^2+1) \text{ combined like terms } \\ \le 2 \cdot 4^k+2(k^3+k^3) \text{ \because } k^3<1 \text{ and } k^3>k^2 \text{ for } k \ge 3 \\ =2 \cdot 4^k+2(2k^3) \text{ combined like terms } \\ \le 2\cdot 4^k+2(4^k) \text{ used induction hypothesis }\]

Answer 48

last step is to basically combine like terms

Answer 49

and use law of exponents

Answer 50

ohh wow thank you very much , this is perfect

Answer 51

you can correct the k^3>1 thing right

Answer 52

i put k^3<1

Answer 53

ok will do thanks again

Answer 54

np