Proving Multiples in Sequence 2, 5, 8, 11, 14

[Natasha1]

Take any number in the sequence

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

and investigate which multiples of that number are in the sequence. Is the result true for all members of the sequence? Prove any result you find.

My answers:

2 gives 2, 8, 14, 20, 26,...
5 gives 5, 20, 35, 50, 65, 80,...
8 gives 8, 32, 56, 80, 104, 128, ...
11 gives 11, 44, 77, 110, 143, 176, ...

It looks as if the any number in the sequence has multiples of that number. Any clever one of you could tell me how to prove this as I am stuck please

 

[Tide]

HINT: Do you think the fact that successive numbers in the original sequence differ by 3 matters? :)

 

[Natasha1]

HINT: Do you think the fact that successive numbers in the original sequence differ by 3 matters? :)

Probably but I am not really more advanced to be honest

The general term is 3n - 1 that much I know

 

[arildno]

How about the following then:
The multiples of 2 obey: 2*1, 2*4, 2*7, 2*10..
The multiples of 5 obey: 5*1, 5*4, 5*7, 5*10..

 

[Natasha1]

How about the following then:
The multiples of 2 obey: 2*1, 2*4, 2*7, 2*10..
The multiples of 5 obey: 5*1, 5*4, 5*7, 5*10..

I see but how from here can I prove it's true for all numbers in the sequence? I need a general solution basically

 

[quasar987]

The key is to find the general term of the sequence of multiple.

You've already found the general term of the original sequence. Call it $a_n$.

Then select any number in the original sequence and find the general term of the sequence of multiples of that number such that the result is also in the original. Call it $b_n$.

Show that when you multiplie an element of $\{a_n\}$ by an element of $\{b_n\}$ the result fits the general form of $a_n$. This means that the result is also in a_n.

 

[Natasha1]

The key is to find the general term of the sequence of multiple.

You've already found the general term of the original sequence. Call it $a_n$.

Then select any number in the original sequence and find the general term of the sequence of multiples of that number such that the result is also in the original. Call it $b_n$.

Show that when you multiplie an element of $\{a_n\}$ by an element of $\{b_n\}$ the result fits the general form of $a_n$. This means that the result is also in a_n.

so the general term of the original sequence is an = 3n + 2

Now, the general term of the sequence of multiples of 2 is 2(3n +2) = 6n + 12 = 6 (n +2) ?

Now what should I do from here?

 

[quasar987]

so the general term of the original sequence is an = 3n + 2

Now, the general term of the sequence of multiples of 2 is 2(3n +2) = 6n + 12 = 6 (n +2) ?

Now what should I do from here?

You said earlier:

The general term is 3n - 1 that much I know

That was correct. If you let n take all values in $\mathbb{N}$, then $\{3n - 1\}_{n\in \mathbb{N}}$ spans all values of the original sequence.

But {2(3n +2)} does not span all values of the sequence of multiples , even if you let n take the value 0 as well as the first number of this sequence is then 4, while you want it to be 1.

The multiples of 2 obey: 2*1, 2*4, 2*7, 2*10..
The multiples of 5 obey: 5*1, 5*4, 5*7, 5*10..

See? the sequence of multiples is 1,4,7,10,...

Now find the correct form of $b_n$ such that $\{b_n\}_{n\in \mathbb{N}} = \{1,4,7,10,...\}$

 

[Natasha1]

You said earlier:



That was correct. If you let n take all values in $\mathbb{N}$, then $\{3n - 1\}_{n\in \mathbb{N}}$ spans all values of the original sequence.

But {2(3n +2)} does not span all values of the sequence of multiples , even if you let n take the value 0 as well as the first number of this sequence is then 4, while you want it to be 1.


See? the sequence of multiples is 1,4,7,10,...

Now find the correct form of $b_n$ such that $\{b_n\}_{n\in \mathbb{N}} = \{1,4,7,10,...\}$

so bn = 3n - 2

What do you mean by an element of {an} or {bn}? Is it a number in the sequence? Sorry I am very new to all this

 

[quasar987]

so bn = 3n - 2

That's better.

What do you mean by an element of {an} or {bn}? Is it a number in the sequence? Sorry I am very new to all this

What I meant is that a number x is an element of {an} if x is a number in the original sequence. A formal way to put it, now that we have the general form of a_n would be to say that x is an element of {an} if $x = 3n_x - 1$ for a certain integer $n_x \in \mathbb{N}$.

Now the idea is to show that any given element of {an} multiplied by any given element of {bn} is in {an}, i.e. is of the form $3n - 1$ for a certain integer $n \in \mathbb{N}$

 

[Natasha1]

That's better.


What I meant is that a number x is an element of {an} if x is a number in the original sequence. A formal way to put it, now that we have the general form of a_n would be to say that x is an element of {an} if $x = 3n_x - 1$ for a certain integer $n_x \in \mathbb{N}$.

Now the idea is to show that any given element of {an} multiplied by any given element of {bn} is in {an}, i.e. is of the form $3n - 1$ for a certain integer $n \in \mathbb{N}$

Ah ha! So if I multiply an * bn = (3n-1)(3n-2) = 9n^2 - 9n +2 ?

 

[quasar987]

How does what you wrote shows that (3n-1)(3n-2) is an element of an ?!?

Start by answering that, it's a good first step. After that we'll look into why just showing this does not solve the problem entirely.

 

[Natasha1]

How does what you wrote shows that (3n-1)(3n-2) is an element of an ?!?

Start by answering that, it's a good first step. After that we'll look into why just showing this does not solve the problem entirely.

The answer is I simply don't know how to do it. I won't lie... my brain just can't take me there as I have never proved anything like this before

 

[quasar987]

Alright, alright, no problem, look. You've found that an * bn = (3n-1)(3n-2) = 9n^2 - 9n +2.

You know that every element of the sequence {am} is of the form 3m - 1, where m in an integer.

To show that an * bn is an element of {am} is then equivalent to showing that an * bn is of the form 3m - 1 where m in an integer.

So, can you algebraically rearange 9n^2 - 9n +2 so it takes on the form 3m - 1?

Hint: +2 = -1 + 3.

 

[Natasha1]

Alright, alright, no problem, look. You've found that an * bn = (3n-1)(3n-2) = 9n^2 - 9n +2.

You know that every element of the sequence {am} is of the form 3m - 1, where m in an integer.

To show that an * bn is an element of {am} is then equivalent to showing that an * bn is of the form 3m - 1 where m in an integer.

So, can you algebraically rearange 9n^2 - 9n +2 so it takes on the form 3m - 1?

Hint: +2 = -1 + 3.

I thought you said the multiplying an*bn was rubbish? So I don't get why you are telling me start from 9n^2 - 9n +2

the answer must therefore be:

9n^2-9n+2 = 3(3n^2-3n+1)-1 = 3K-1 which is of the form am

 

[quasar987]

I thought you said the multiplying an*bn was rubbish?

When did I say that? I said it was a good first step.

the answer must therefore be:

9n^2-9n+2 = 3(3n^2-3n+1)-1 = 3K-1 which is of the form am

Correct!

Now the only problem is that you only proved that the nth term of thhe sequence {am} times the nth term of the sequence {bm} is in {am}. You need to show that any elment of {am} times any element of {bm} was in {am}.

To do this, simply show, as you just did, that $a_n_1 * b_n_2 = (3n_1 - 1)(3n_2 - 2)$ is in {am}.

 

[Natasha1]

When did I say that? I said it was a good first step.



Correct!

Now the only problem is that you only proved that the nth term of thhe sequence {am} times the nth term of the sequence {bm} is in {am}. You need to show that any elment of {am} times any element of {bm} was in {am}.

To do this, simply show, as you just did, that $a_n_1 * b_n_2 = (3n_1 - 1)(3n_2 - 2)$ is in {am}.

I get at the end 3(3n1n2 - 2n1 - n2 +1) - 1 = 3K-1 is that it?

 

[quasar987]

I get at the end 3(3n1n2 - 2n1 - n2 +1) - 1 = 3K-1 is that it?

Well is 3n1n2 - 2n1 - n2 +1 a positive integer? If so, then 3(3n1n2 - 2n1 - n2 +1) - 1 is in am, and you've won. You've won the privilege of writing your very first

$$\mathcal{Q.E.D.}$$

Bravo!

 

[Natasha1]

Well is 3n1n2 - 2n1 - n2 +1 a positive integer? If so, then 3(3n1n2 - 2n1 - n2 +1) - 1 is in am, and you've won. You've won the privilege of writing your very first

$$\mathcal{Q.E.D.}$$

Bravo!

How do you prove it's a positive integer?

 

[quasar987]

How do you prove it's a positive integer?

we found that $(3n_1 -1)(3n_2 - 2) = 3K -1$ and now we want to show that $K\geq 1$. Well first we notice that $3n_1 -1 \geq 2$ and $3n_2 -2\geq 1$. So $(3n_1 -1)(3n_2 - 2)\geq 2$ Hence, $3K -1 \geq 2 \Leftrightarrow 3K \geq 3 \Leftrightarrow K \geq 1$

 

[VietDao29]

we found that $(3n_1 -1)(3n_2 - 2) = 3K -1$ and now we want to show that $K\geq 1$. Well first we notice that $3n_1 -1 \geq 2$ and $3n_2 -2\geq 1$. So $(3n_1 -1)(3n_2 - 2)\geq 2$ Hence, $3K -1 \geq 2 \Leftrightarrow 3K \geq 3 \Leftrightarrow K \geq 1$

quasar987, by this way you have proved that a_n b_m must be an element of {a_n}. But, what if there exists some number x, and some number i such that: a_x i is also an element of {a_n}, but i is not an element {b_m}?
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May I suggest a little different way:
$$a_n \equiv 2 \mbox{ mod } 3, \ \forall n \in \mathbb{N} ^ *$$
If a_n * b_m must also in {a_n}, that means:
$$a_n \ b_m \equiv 2 \mbox{ mod } 3, \ \forall n, m \in \mathbb{N} ^ *$$
So what's $$b_m \equiv ? \mbox{ mod } 3$$?

 

[quasar987]

that's an elegant and more effective way!