Mathematical induction problem

[DDarthVader]

Homework Statement

Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(

Homework Equations

The Attempt at a Solution

Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?

 

[professordad]

You're only working with odd numbers, so the next term wouldn't be $2n$..

 

[DDarthVader]

This is what I should do to solve the problem $1+3+5+...+(2n-1+1)=1+3+5+...+2n$. Right? Because you always do $n+1$ in the Induction step.

 

[sacscale]

Homework Statement

Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(

Homework Equations

The Attempt at a Solution

Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?

Mathematical induction attempts to show that if the equation holds for n, then it also holds for n+1. It's easier to keep straight if you use a substitution such as n=k+1

 

[professordad]

@ DDarthVader: Yes, but you're not substituting $n + 1$ in for $2n - 1$ or anything, you're subbing it in for $n$...

 

[DDarthVader]

Doing the substitution $n=k+1$ I've got this result:
$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$
Then
$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$
But since $n=k+1$ we got
$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$
Is this correct?

 

[SammyS]

Your conjecture: $1+3+5+...+(2n-1)=n^2$

You've already done the base step, n = 1.

For the induction step:

Let k ≥ 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .

So you need to show that $1+3+5+\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,

starting with $1+3+5+\dots+(2k-1)=k^2\ .$

 

[azizlwl]

Homework Statement

Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(

Homework Equations

The Attempt at a Solution

Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?

As you see here, (n+1)^th value is not equal to 2n
It is equal to (2(n+1)-1)

 

[sacscale]

Doing the substitution $n=k+1$ I've got this result:
$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$
Then
$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$
But since $n=k+1$ we got
$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$
Is this correct?

Substituting n=k+1 into 2n-1 gives 2(k+1)-1.

 

[DDarthVader]

Your conjecture: $1+3+5+...+(2n-1)=n^2$

You've already done the base step, n = 1.

For the induction step:

Let k ≥ 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .

So you need to show that $1+3+5+\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,

starting with $1+3+5+\dots+(2k-1)=k^2\ .$

Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?

 

[Villyer]

Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?

Isn't it bad practice to manipulate both sides of an equation in a problem such as this?

 

[SammyS]

Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?

Yes, somewhat indirectly.

Rather you should do something like the following.

Assume $1+3+5+\dots+(2k-1)=k^2\ .$

Now consider:

$1+3+5+\dots+(2k-1)+(2(k+1)-1)$

$=k^2+(2(k+1)-1)$   ... because of or assumption.

$=k^2+2k+1$

$=(k+1)^2$  Which is what we needed to show for the inductive step.​

It's not that what you did is particularly wrong, it's just that it's not real clear that you're not somehow assuming the very thing you should be proving.

 

[Muphrid]

It's generally frowned upon to work both sides at once, yeah. Usually, one would take $k^2 + 2(k+1)-1$ and manipulate it until it's clear the result is $(k+1)^2$. Working only in one direction ensures that no illegal operations or cancellations are done, even though it may be necessary to set both sides equal just to figure out how to do it.

 

[DDarthVader]

Well, I think I got it now! I have a list of mathematical induction to do. I'll try to solve the exercises using what you guys told me. I'll probably be back soon. Thanks guys!

 

[amrah]

Prove by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n.
Please help me in solving this question!

 

[Infinitum]

Prove by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n.
Please help me in solving this question!

Please create a new thread for this.

Also, show how you approached it so we can help you better!

As a start...try showing it is true for n=1..then assume it to be true for n=k...