Mathematical Induction Problem (1 Viewer)

[Tryingtodowell]

Can someone check this Im not sure for part A :
For part b - I literally cant do it Idk how to approach it

A) Use mathematical Induction to prove that 3^n >n^3 for all integers n >=4

For n=4
LHS= 81
RHS= 64

Therefore LHS > RHS
Thus true

Assume true for n=k, k all real

3^k > k^3

RTP; n=k+1

3^k+1 > k+1^3

LHS= 3(3k)
> 3(K^3) from assumption

3(K^3) > (k+1)^3 since 3k > k+1 as n>=4

Therefore through P of Mi....

Part B) Hence or otherwise show that cube root 3 > n root n for all integers n>=4

I dont know how to do this hence question following up from part a ( and since it was given very little space compared to part A I dont think u go through the induction procedure again so how would u approach this)

Thanks

 

[liamkk112]

Can someone check this Im not sure for part A :
For part b - I literally cant do it Idk how to approach it

A) Use mathematical Induction to prove that 3^n >n^3 for all integers n >=4

For n=4
LHS= 81
RHS= 64

Therefore LHS > RHS
Thus true

Assume true for n=k, k all real

3^k > k^3

RTP; n=k+1

3^k+1 > k+1^3

LHS= 3(3k)
> 3(K^3) from assumption

3(K^3) > (k+1)^3 since 3k > k+1 as n>=4

Therefore through P of Mi....

Part B) Hence or otherwise show that cube root 3 > n root n for all integers n>=4

I dont know how to do this hence question following up from part a ( and since it was given very little space compared to part A I dont think u go through the induction procedure again so how would u approach this)

Thanks

(from part a)


then raising both sides to 1/n:

or that

 

[Tryingtodowell]

(from part a)


then raising both sides to 1/n:

or that

???

 

[liamkk112]

???

do u need me to clarify something

 

[Tryingtodowell]

do u need me to clarify something

It doesnt pop up

I see
(from part A)

then raising both sides to

or that

 

[Tryingtodowell]

Oh nvm it did sorry

 

[Tryingtodowell]

(from part a)


then raising both sides to 1/n:

or that

Is part a fine tho?

 

[Luukas.2]

​

 

[Tryingtodowell]

​

I dont see anything... its blank for me

 

[Luukas.2]

I dont see anything... its blank for me

Try reloading the page...

 

[Tryingtodowell]

Try reloading the page...

I did a billion times mate finally works now tho

 

[Luukas.2]

Having got to LHS > 3k³, you could then consider the graph of y = 3x³ - (x + 1)³ = 2x³ - 3x² - 3x + 1, which is a cubic with both of its stationary points below the x-axis and a single x-intercept (a little above x = 2) and so y > 0 for all x > 3, which means LHS > RHS.

 

[Sethio]

Can someone check this Im not sure for part A :
For part b - I literally cant do it Idk how to approach it

A) Use mathematical Induction to prove that 3^n >n^3 for all integers n >=4

For n=4
LHS= 81
RHS= 64

Therefore LHS > RHS
Thus true

Assume true for n=k, k all real

3^k > k^3

RTP; n=k+1

3^k+1 > k+1^3

LHS= 3(3k)
> 3(K^3) from assumption

3(K^3) > (k+1)^3 since 3k > k+1 as n>=4

Therefore through P of Mi....

Part B) Hence or otherwise show that cube root 3 > n root n for all integers n>=4

I dont know how to do this hence question following up from part a ( and since it was given very little space compared to part A I dont think u go through the induction procedure again so how would u approach this)

Thanks

yo hol up that's my dr du question. that's crazy

 

[WeiWeiMan]

yo hol up that's my dr du question. that's crazy

that's crazy bro