Question

Prove that 4n 11 years ago

Answer 1

induction maybe works

Answer 2

Is an induction problem I forgot to put that.

Answer 3

I assumed you have no trouble with the base case?

Answer 4

no

Answer 5

So you know to suppose
\[4k<k! \text{ is true for integer } k>8\]
and show
\[4(k+1)<(k+1)! \]

Answer 6

I can't seem to get the first part \[4^{k+1}=4\times \] , but i know the others

Answer 7

< (k+1) ⋅k!= (k+1)!,
which proves the statement for n= k+1

Answer 8

oh is that an exponent ?

Answer 9

it is 4^k<k! ?

Answer 10

\[4^k <k! \\ \text{ try multiplying both sides by (k+1)} \\ 4^k(k+1)<(k+1)!\]
hint k+1>4 for all integer k>8

Answer 11

that is...
\[?<4^k(k+1)<(k+1)!\]

Answer 12

what should be in place of that question

Answer 13

what do we want to show
what can we show with this