# [SOLVED]Prove with mathematical induction

#### wishmaster

##### Active member
With mathematical induction i should prove that is true for all natural numbers:

Im sorry beacuse i have inserted an image,but im still not used to write it in $$\displaystyle here.....$$

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#### MarkFL

Staff member
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?

#### wishmaster

##### Active member
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 thats true if N = 1

#### MarkFL

Staff member
12*(1+1)2/4 = 1
So i believe thats 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.

#### wishmaster

##### Active member
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

Staff member
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.

#### wishmaster

##### Active member
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

Staff member
$$\displaystyle 1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$ ??
Yes, that's right!

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...

#### wishmaster

##### Active member
Yes, that's right!

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? Im off the work,dont have time now...will come at the evening! Thank you again!

#### wishmaster

##### Active member
Will try that,thank you!

#### MarkFL

Staff member
Can u show me please? Im 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?

#### wishmaster

##### Active member
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

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

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 $(k+1)^2$.

#### wishmaster

##### Active member
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

Staff member
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.

#### wishmaster

##### Active member
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!