Question

Help with another proof
http://prntscr.com/3h1sre

Answer 1

i can see induction here use it twice ,
first by induction prove that
n^3>n^2

Answer 2

\(n=2 : \\ 2^3 > 2^2 \implies 2 \in S\)

assume : \(k^3 > k^2\)

\(n = k+1 : \\ (k+1)^3 = k^3 + 3k^2 + 3k + 1 = k^3 + 2k^2 + k + (k+1)^2 > (k+1)^2\)

\(\implies n^3 \gt n^2\) for all \(n \ge 2\)

Answer 3

@BSwan next what

Answer 4

\(n = 4 : 4! > 4^2 \implies n = 4 \in S\)

assume \(k! > k^2\)

\(k+1 : \\ k!(k+1) \gt k^2(k+1) \implies (k+1)! \gt k^3 + k^2 \)
.... ??

Answer 5

I dont seem to able to conclude neatly.. :(

Answer 6

a hint would be sufficient from here..

Answer 7

I was never good with inequalities for induction so im not too sure if im doing it right not

Assume
\[k! > k^2\]
and prove for k+1
\[(k+1)! > (k+1)^2\]

\[(k+1)! = k!(k+1) > (k+1)k^2 > k^2 + 2k + 1\]

Answer 8

yeah \((k+1)k^2 = k^3 + k^2 > k^2 + 2k + 1\) ? another induction sub proof ?

Answer 9

i was thinking that k^3 should be > 2k+1 by looking at it since we are looking at values greater than or equal to 4

but we might need to do another induction on that probably

Answer 10

i think so we may use k^3 > 2k+1 and conclude thank you @jayz657 :)

Answer 11

alright np then glad to help

and the trasition from k!(k+1) > (k+1)k^2 is by Inductive Hypothesis, just forgot to metion it

Answer 12

yeah got that :

k! > k^2

=>

k!(k+1) > (k+1)k^2

Answer 13

il move to other questions thanks again :D

Answer 14

alright then gl ^^

Answer 15

@ganeshie8 message me back when you get a chance please :)