Prove with mathematical induction

[theakdad]

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 \(\displaystyle here...\)

 

[MarkFL]

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?

 

[theakdad]

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

 

[MarkFL]

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

I would actually write:

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

\(\displaystyle 1=1\)

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.

 

[theakdad]

I would actually write:

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

\(\displaystyle 1=1\)

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

 

[MarkFL]

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:

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

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.

 

[theakdad]

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:

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

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.

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

 

[MarkFL]

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

Yes, that's right! :D

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

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

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

 

[theakdad]

Yes, that's right! :D

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

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

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!

 

[Petrus]

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,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

Will try that,thank you!

 

[MarkFL]

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 \(\displaystyle \frac{(k+1)^2}{4}\). What do you get?

 

[theakdad]

Try factoring out \(\displaystyle \frac{(k+1)^2}{4}\). What do you get?

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

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?

 

[MarkFL]

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

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 $(k+1)^2$.

 

[theakdad]

Try factoring that further...the two terms in the numerator have a common factor of $(k+1)^2$.

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

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

 

[MarkFL]

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

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:

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

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.

 

[theakdad]

Yes, and we can now write:

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

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!