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[SOLVED] Prove with mathematical induction

wishmaster

Active member
Oct 11, 2013
211
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 \(\displaystyle here.....\)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle 1^3+2^3+3^3+\cdots+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3\) ??
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...
 

wishmaster

Active member
Oct 11, 2013
211
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? Im off the work,dont have time now...will come at the evening! Thank you again!
 

Petrus

Well-known member
Feb 21, 2013
739

wishmaster

Active member
Oct 11, 2013
211
Will try that,thank you!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{k^2(k+1)^2+4(k+1)^3}{4}\)

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
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
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!