Mathematical Induction Inequality (2 Viewers)

[Tryingtodowell]

Hello I just joined

Im struggling with mathematical induction inequality so bad and someone recommended me here

Need help with these:

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
2. Prove by mathematical induction that 3^n > 1+2n for all integers n>2
3.Prove by induction that 12^n > 7^n +5^n for integers n>2

and more but I want to understand the technique cause I just cant seem to know how to write the 'meat' of the answer

yes its basic cause I just started the topic dont judge me

 

[Tryingtodowell]

plz someone

 

[cossine]

and more but I want to understand the technique cause I just cant seem to know how to write the 'meat' of the answer

Prove the base case.
Prove that if the statement is true for n = k then it true for n = k+1 #(1)
Repeatly apply (1) to prove it is true for all n.

For some question you may need to make use lemmas.

It would be good if you made an attempt first and presented your attempted solution

 

[Tryingtodowell]

Prove the base case.
Prove that if the statement is true for n = k then it true for n = k+1 #(1)
Repeatly apply (1) to prove it is true for all n.

For some question you may need to make use lemmas.

It would be good if you made an attempt first and presented your attempted solution

My main problem is how to go about the question after letting n=k and n=k+1

 

[cossine]

My main problem is how to go about the question after letting n=k and n=k+1

Yes but surely you must have attempted something. You need to somehow make use of the statement being true for n= k.

It would be good if you used a textbook with solutions

 

[Average Boreduser]

My main problem is how to go about the question after letting n=k and n=k+1

show some working. no point in us making a whole khan academy lecture if you're only missing 1 or 2 things

 

[Tryingtodowell]

show some working. no point in us making a whole khan academy lecture if you're only missing 1 or 2 things

No Im missing the rest I can easily say Let n=k and n=K+1 for the given expression but I cant do the rest

 

[WeiWeiMan]

No Im missing the rest I can easily say Let n=k and n=K+1 for the given expression but I cant do the rest

basically these sorta questions generally have you replace someone like replacing k^2 with k (since k^2 > k for the given domain)or stuff like that
watch the mcgrathematics video on it

 

[Tryingtodowell]

Hello I just joined

Im struggling with mathematical induction inequality so bad and someone recommended me here

Need help with these:

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
2. Prove by mathematical induction that 3^n > 1+2n for all integers n>2
3.Prove by induction that 12^n > 7^n +5^n for integers n>2

and more but I want to understand the technique cause I just cant seem to know how to write the 'meat' of the answer

yes its basic cause I just started the topic dont judge me

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
For n=1
LHS= (2)^2+1= 5
RHS=2

Thus LHS>RHS, therefore true

Assume n=k,
k^2+1 >K

RTP; n=k+1
(k+1)^2 +1 > k+1

Now what do I do and how do I go about it

 

[Tryingtodowell]

basically these sorta questions generally have you replace someone like replacing k^2 with k (since k^2 > k for the given domain)or stuff like that
watch the mcgrathematics video on it

Have already spent hours trynna find yt vids and stuff to help me out but im so lost pls help

 

[Tryingtodowell]

Yes but surely you must have attempted something. You need to somehow make use of the statement being true for n= k.

It would be good if you used a textbook with solutions

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
For n=1
LHS= (2)^2+1= 5
RHS=2

Thus LHS>RHS, therefore true

Assume n=k,
k^2+1 >K

RTP; n=k+1
(k+1)^2 +1 > k+1

Now what do I do and how do I go about it

 

[WeiWeiMan]

Have already spent hours trynna find yt vids and stuff to help me out but im so lost pls help

bruv watch mcgrathematics

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
For n=1
LHS= (2)^2+1= 5
RHS=2

Thus LHS>RHS, therefore true

Assume n=k,
k^2+1 >K

RTP; n=k+1
(k+1)^2 +1 > k+1

Now what do I do and how do I go about it

expand it and use assumption

 

[liamkk112]

Hello I just joined

Im struggling with mathematical induction inequality so bad and someone recommended me here

Need help with these:

1. Using Mathematical induction, prove that n^2 + 1 > n for all integers n>1
2. Prove by mathematical induction that 3^n > 1+2n for all integers n>2
3.Prove by induction that 12^n > 7^n +5^n for integers n>2

and more but I want to understand the technique cause I just cant seem to know how to write the 'meat' of the answer

yes its basic cause I just started the topic dont judge me

for 1, base case is pretty obvious: we let n = 2, then n^2 + 1 = 4 + 1 = 5, and n= 2, 5>2 so base case is true

our assumption is assuming that the statement is true for n=k, k^2 +1 > k for all integers k > 1

we now prove that the statement is true for n = k +1, or that (k+ 1)^2 +1 > k +1

LHS = (k+1)^2 + 1 = k^2 +2k + 1 + 1 = (k^2 + 1) + 2k + 1
> k + 2k + 1 by our assumption
= 3k + 1
> k + 1 since k >1, so 3k>k
= RHS

hence LHS>RHS as required, then by mathematical induction...

similar steps for the rest

 

[liamkk112]

basically you want to get the k+1 part into the form of the assumption, so u can apply the assumed inequality

 

[Tryingtodowell]

for 1, base case is pretty obvious: we let n = 2, then n^2 + 1 = 4 + 1 = 5, and n= 2, 5>2 so base case is true

our assumption is assuming that the statement is true for n=k, k^2 +1 > k for all integers k > 1

we now prove that the statement is true for n = k +1, or that (k+ 1)^2 +1 > k +1

LHS = (k+1)^2 + 1 = k^2 +2k + 1 + 1 = (k^2 + 1) + 2k + 1
> k + 2k + 1 by our assumption
= 3k + 1
> k + 1 since k >1, so 3k>k
= RHS

hence LHS>RHS as required, then by mathematical induction...

similar steps for the rest

thank u so much

 

[Tryingtodowell]

I kinda get it now thx

 

[Tryingtodowell]

The base case seems a bit weird, You have n=1 for base case but require n>1 (not that it matters)

assumption is good.

LHS = (k+1)^2 +1
= k^2 + 2k + 1 + 1
>= 3k + 2 (using assumption)
>= 3k + 1
> k + 1 (since 3k > k since k > 0)

Therefore the statement is true by mathematical induction.

thankks yes because its meant to be n>/1

 

[Tryingtodowell]

2. Prove by mathematical induction that 3^n > 1+2n for all integers n>/2

For n=2
LHS= 9
RHS=1+2(2)=5

Thus LHS>RHS
True

Assume n=k
3^k >1+2k

RTP; n=k+1
3^k+1 > 1+2(k+1)
3^k+1> 2k+3

LHS= 3(1+2k)
= 3+6k (using assumption)
>2k+3
Thus, 3^k+1>2k+3

Therefore true for n=1, k, k+1

Thus n>1,2,3...

Therefore by P of Mi 3^n>1+2n for n>/2

Can someone check this because Im not confident the assumptions bit and generally overall

 

[liamkk112]

2. Prove by mathematical induction that 3^n > 1+2n for all integers n>/2

For n=2
LHS= 9
RHS=1+2(2)=5

Thus LHS>RHS
True

Assume n=k
3^k >1+2k

RTP; n=k+1
3^k+1 > 1+2(k+1)
3^k+1> 2k+3

LHS= 3(1+2k)
= 3+6k (using assumption)
>2k+3
Thus, 3^k+1>2k+3

Therefore true for n=1, k, k+1

Thus n>1,2,3...

Therefore by P of Mi 3^n>1+2n for n>/2

Can someone check this because Im not confident the assumptions bit and generally overall

um im not sure about what you did for the n = k + 1 but it should be something like this (starting off from n = k +1 proof):

LHS = 3^(k+1) = 3 . 3^k
> 3 . (1+2k) by assumption
= 6k + 3
> 2k + 3 since 6k >2k as k >= 2
= RHS

hence by mathematical induction ...

i think maybe you miswrote 3^(k+1) as (3^k )+ 1, but the +1 should also be in the exponent

 

[Tryingtodowell]

um im not sure about what you did for the n = k + 1 but it should be something like this (starting off from n = k +1 proof):

LHS = 3^(k+1) = 3 . 3^k
> 3 . (1+2k) by assumption
= 6k + 3
> 2k + 3 since 6k >2k as k >= 2
= RHS

hence by mathematical induction ...

i think maybe you miswrote 3^(k+1) as (3^k )+ 1, but the +1 should also be in the exponent

ok