Prove with mathematical induction

AI Thread Summary
The discussion focuses on proving a mathematical statement using induction, starting with the base case for n=1, which is confirmed as true. Participants outline the process of establishing the induction hypothesis and deriving the case for n+1, emphasizing the importance of factoring in the inductive step. The correct formulation for the hypothesis is presented, and the proof is completed by showing that both sides of the equation are equivalent after factoring. The thread concludes with a successful confirmation that the proof by induction is valid for all natural numbers.
theakdad
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With mathematical induction i should prove that is true for all natural numbers:http://img.tapatalk.com/d/13/10/18/u6epesu5.jpg
Im sorry beacuse i have inserted an image,but I am still not used to write it in $$ here...
 
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The first thing you want to do is show that the base case is true, that is, to show that it holds for $n=1$. Can you demonstrate this?
 
MarkFL said:
The first thing you want to do is show that the base case is true, that is, to show that it holds for $n=1$. Can you demonstrate this?

12*(1+1)2/4 = 1
So i believe that's true if N = 1
 
andreask said:
12*(1+1)2/4 = 1
So i believe that's true if N = 1

I would actually write:

$$1^3=\frac{1^2(1+1)^2}{4}$$

$$1=1$$

So, yes, the base case $P_1$ is true. So next, state the induction hypothesis $P_k$. Just replace $n$ with $k$ in what you are trying to prove.
 
MarkFL said:
I would actually write:

$$1^3=\frac{1^2(1+1)^2}{4}$$

$$1=1$$

So, yes, the base case $P_1$ is true. So next, state the induction hypothesis $P_k$. Just replace $n$ with $k$ in what you are trying to prove.

I have to prove it for n+1
 
andreask said:
I have to prove it for n+1

You want to prove it is true for all $n\in\mathbb{N}$, that is, for all natural numbers $n$. So, state the induction hypothesis $P_k$...and then see if you can derive $P_{k+1}$ by adding what is needed to both sides, this is called the inductive step.

The induction hypothesis $P_{k}$ is:

$$1^3+2^3+3^3+\cdots+k^3=\frac{k^2(k+1)^2}{4}$$

Now, what do you suppose a good inductive step would be? What would need to be added to the left side so that it would correspond to $P_{k+1}$? Once you decide what this is, then add it to both sides of $P_{k}$, and then see if the right side can be written as it is for $P_{k}$, but with $k$ replaced by $k+1$.

Once you do this, then you will have derived $P_{k+1}$ from $P_{k}$ and thereby completed the proof by induction.
 
MarkFL said:
You want to prove it is true for all $n\in\mathbb{N}$, that is, for all natural numbers $n$. So, state the induction hypothesis $P_k$...and then see if you can derive $P_{k+1}$ by adding what is needed to both sides, this is called the inductive step.

The induction hypothesis $P_{k}$ is:

$$1^3+2^3+3^3+\cdots+k^3=\frac{k^2(k+1)^2}{4}$$

Now, what do you suppose a good inductive step would be? What would need to be added to the left side so that it would correspond to $P_{k+1}$? Once you decide what this is, then add it to both sides of $P_{k}$, and then see if the right side can be written as it is for $P_{k}$, but with $k$ replaced by $k+1$.

Once you do this, then you will have derived $P_{k+1}$ from $P_{k}$ and thereby completed the proof by induction.
$$1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$ ??
 
andreask said:
$$1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$ ??

Yes, that's right! :D

So, you want to see if the right side can be written as:

$$\frac{(k+1)^2((k+1)+1)^2}{4}$$

As you can see, factoring would be a good first step...
 
MarkFL said:
Yes, that's right! :D

So, you want to see if the right side can be written as:

$$\frac{(k+1)^2((k+1)+1)^2}{4}$$

As you can see, factoring would be a good first step...

Can u show me please? I am off the work,dont have time now...will come at the evening! Thank you again!
 
  • #10
andreask said:
Can u show me please? I am off the work,dont have time now...will come at the evening! Thank you again!
Hello,
This explain how induction proof works Proof by Induction | Induction | Khan Academy
Regards,
$$|\pi\rangle$$
 
  • #11
Will try that,thank you!
 
  • #12
andreask said:
Can u show me please? I am off the work,dont have time now...will come at the evening! Thank you again!

Try factoring out $$\frac{(k+1)^2}{4}$$. What do you get?
 
  • #13
MarkFL said:
Try factoring out $$\frac{(k+1)^2}{4}$$. What do you get?

$$\frac{k^2(k+1)^2+4(k+1)^3}{4}$$

Now i think i got it..i know where my problem is,im not good in factoring out,any ideas how can i become better?
 
  • #14
andreask said:
$$\frac{k^2(k+1)^2+4(k+1)^3}{4}$$

Now i think i got it..i know where my problem is,im not good in factoring out,any ideas how can i become better?

Try factoring that further...the two terms in the numerator have a common factor of $(k+1)^2$.
 
  • #15
MarkFL said:
Try factoring that further...the two terms in the numerator have a common factor of $(k+1)^2$.

Its $$\frac{(k+1)^2(k+2)^2}{4}$$

So the left and right side are the same, that proves that P is good for all natural numbers,right?
 
  • #16
andreask said:
Its $$\frac{(k+1)^2(k+2)^2}{4}$$

So the left and right side are the same, that proves that P is good for all natural numbers,right?

Yes, and we can now write:

$$1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{(k+1)^2((k+1)+1)^2}{4}$$

I wrote the right side that way to make it clear that it now has $k+1$ everywhere that $P_{k}$ has $k$.

Hence, this is $P_{k+1}$, which we derived from $P_{k}$, and so we have completed the proof by induction.
 
  • #17
MarkFL said:
Yes, and we can now write:

$$1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{(k+1)^2((k+1)+1)^2}{4}$$

I wrote the right side that way to make it clear that it now has $k+1$ everywhere that $P_{k}$ has $k$.

Hence, this is $P_{k+1}$, which we derived from $P_{k}$, and so we have completed the proof by induction.

Thank you very much!
 
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