# 2, 5, 8, 11, 14

1. Feb 12, 2006

### 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.

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

Last edited: Feb 12, 2006
2. Feb 12, 2006

### Tide

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

3. Feb 12, 2006

### Natasha1

Probably but I am not really more advanced to be honest

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

Last edited: Feb 12, 2006
4. Feb 12, 2006

### arildno

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..

5. Feb 12, 2006

### Natasha1

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

6. Feb 12, 2006

### 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.

7. Feb 12, 2006

### Natasha1

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?

8. Feb 12, 2006

### quasar987

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 $\{b_n\}_{n\in \mathbb{N}} = \{1,4,7,10,...\}$

9. Feb 12, 2006

### Natasha1

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

Last edited: Feb 12, 2006
10. Feb 12, 2006

### quasar987

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 $n \in \mathbb{N}$

Last edited: Feb 12, 2006
11. Feb 12, 2006

### Natasha1

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

12. Feb 12, 2006

### 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.

13. Feb 12, 2006

### Natasha1

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

14. Feb 12, 2006

### 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.

15. Feb 12, 2006

### Natasha1

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

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

16. Feb 12, 2006

### quasar987

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}.

17. Feb 12, 2006

### Natasha1

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

18. Feb 12, 2006

### quasar987

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!

Last edited: Feb 12, 2006
19. Feb 12, 2006

### Natasha1

How do you prove it's a positive integer?

20. Feb 12, 2006

### quasar987

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, $3K -1 \geq 2 \Leftrightarrow 3K \geq 3 \Leftrightarrow K \geq 1$