2 5 8 11 14
2, five, 8, 11, fourteen
- Thread starter Natasha1
- Offset date
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 observe.
My answers:
2 gives 2, 8, 14, twenty, 26,...
five gives 5, 20, 35, 50, 65, eighty,...
8 gives eight, 32, 56, 80, 104, 128, ...
eleven gives 11, 44, 77, 110, 143, 176, ...
It looks equally if the any number in the sequence has multiples of that number. Whatever clever one of you could tell me how to evidence this equally I am stuck delight 
Answers and Replies
HINT: Do you retrieve the fact that successive numbers in the original sequence differ by iii matters? :)
Probably simply I am not really more advanced to be honest
The general term is 3n - 1 that much I know
The multiples of 2 obey: 2*1, ii*four, ii*vii, 2*ten..
The multiples of 5 obey: 5*1, five*4, 5*seven, five*ten..
How most the post-obit then:
The multiples of ii obey: 2*1, ii*4, 2*7, 2*10..
The multiples of 5 obey: five*1, 5*4, five*vii, 5*10..
I see
but how from hither can I prove it'due south true for all numbers in the sequence? I need a general solution basicallyYou've already institute the general term of the original sequence. Phone call information technology [itex]a_n[/itex].
Then select any number in the original sequence and discover the general term of the sequence of multiples of that number such that the result is also in the original. Call it [itex]b_n[/itex].
Show that when you multiplie an element of [itex]\{a_n\}[/itex] by an element of [itex]\{b_n\}[/itex] the result fits the general form of [itex]a_n[/itex]. This means that the result is as well in a_n.
The key is to find the full general term of the sequence of multiple.You've already found the general term of the original sequence. Call it [itex]a_n[/itex].
Then select any number in the original sequence and find the general term of the sequence of multiples of that number such that the consequence is too in the original. Call it [itex]b_n[/itex].
Show that when you lot multiplie an element of [itex]\{a_n\}[/itex] past an chemical element of [itex]\{b_n\}[/itex] the result fits the general form of [itex]a_n[/itex]. This means that the upshot is also in a_n.
so the general term of the original sequence is an = 3n + ii
Now, the general term of the sequence of multiples of 2 is 2(3n +2) = 6n + 12 = 6 (n +2) ?
Now what should I practice from here?
so the general term of the original sequence is an = 3n + 2Now, the full general term of the sequence of multiples of 2 is 2(3n +ii) = 6n + 12 = vi (n +2)
Now what should I practice from here?
You lot said before:
The general term is 3n - one that much I know
That was correct. If you let n have all values in [itex]\mathbb{N}[/itex], and so [itex]\{3n - 1\}_{due north\in \mathbb{N}}[/itex] spans all values of the original sequence.
But {ii(3n +2)} does non bridge all values of the sequence of multiples , even if you permit northward take the value 0 as well as the first number of this sequence is then four, while you want it to exist ane.
See? the sequence of multiples is ane,four,7,10,...The multiples of ii obey: ii*one, 2*4, two*seven, 2*x..
The multiples of 5 obey: 5*ane, five*4, five*7, 5*x..
Now find the correct class of [itex]b_n[/itex] such that [itex]\{b_n\}_{northward\in \mathbb{N}} = \{i,4,7,x,...\}[/itex]
You said earlier:That was correct. If you let n take all values in [itex]\mathbb{North}[/itex], then [itex]\{3n - 1\}_{due north\in \mathbb{Northward}}[/itex] spans all values of the original sequence.
But {2(3n +2)} does non span all values of the sequence of multiples , even if you let n have the value 0 too every bit the first number of this sequence is so 4, while you desire it to be 1.
Come across? the sequence of multiples is 1,4,vii,ten,...
Now find the correct form of [itex]b_n[/itex] such that [itex]\{b_n\}_{n\in \mathbb{N}} = \{1,4,7,x,...\}[/itex]
so bn = 3n - 2
What do you lot mean by an chemical element of {an} or {bn}? Is it a number in the sequence? Sorry I am very new to all this
That's meliorate.
What I meant is that a number x is an element of {an} if x is a number in the original sequence. A formal mode to put it, at present that we have the general form of a_n would be to say that x is an element of {an} if [itex]10 = 3n_x - one[/itex] for a certain integer [itex]n_x \in \mathbb{Due north}[/itex].What exercise you mean by an element of {an} or {bn}? Is it a number in the sequence? Sorry I am very new to all this
Now the idea is to bear witness that any given element of {an} multiplied by any given element of {bn} is in {an}, i.e. is of the form [itex]3n - 1[/itex] for a certain integer [itex]north \in \mathbb{N}[/itex]
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 style to put information technology, now that we have the full general form of a_n would be to say that x is an element of {an} if [itex]ten = 3n_x - one[/itex] for a certain integer [itex]n_x \in \mathbb{N}[/itex].
At present 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 [itex]3n - i[/itex] for a certain integer [itex]n \in \mathbb{Northward}[/itex]
Ah ha! So if I multiply an * bn = (3n-one)(3n-two) = 9n^2 - 9n +2 ???
Start by answering that, information technology's a adept first footstep. Subsequently that we'll wait into why just showing this does not solve the trouble entirely.
How does what you wrote shows that (3n-i)(3n-ii) is an element of an ?!?Start past answering that, it's a proficient first step. After that we'll expect into why just showing this does not solve the problem entirely.
The answer is I only don't know how to do it. I won't prevarication... my brain merely can't accept me at that place as I have never proved anything like this before
You know that every chemical element of the sequence {am} is of the form 3m - 1, where thousand in an integer.
To show that an * bn is an element of {am} is then equivalent to showing that an * bn is of the class 3m - 1 where m in an integer.
And then, can y'all algebraically rearange 9n^2 - 9n +2 and then it takes on the form 3m - 1?
Hint: +two = -1 + 3.
Alright, alright, no problem, expect. Yous've constitute that an * bn = (3n-one)(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 so equivalent to showing that an * bn is of the class 3m - one where m in an integer.
So, can you algebraically rearange 9n^ii - 9n +two so it takes on the form 3m - 1?
Hint: +2 = -ane + three.
I idea you said the multiplying an*bn was rubbish??? Then I don't get why you are telling me start from 9n^2 - 9n +ii
the answer must therefore exist:
9n^2-9n+ii = iii(3n^2-3n+1)-1 = 3K-1 which is of the course am
When did I say that? I said information technology was a skilful first step.I thought you said the multiplying an*bn was rubbish???
the respond must therefore be:9n^2-9n+ii = three(3n^two-3n+1)-one = 3K-1 which is of the form am
Right!
Now the only trouble 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 lot need to show that any elment of {am} times any element of {bm} was in {am}.
To practise this, simply bear witness, as you but did, that [itex]a_n_1 * b_n_2 = (3n_1 - 1)(3n_2 - 2)[/itex] is in {am}.
When did I say that? I said it was a good outset step.Correct!
Now the only problem is that you just 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 practise this, merely prove, equally yous only did, that [itex]a_n_1 * b_n_2 = (3n_1 - 1)(3n_2 - ii)[/itex] is in {am}.
I get at the cease 3(3n1n2 - 2n1 - n2 +1) - 1 = 3K-one is that information technology?
I get at the terminate 3(3n1n2 - 2n1 - n2 +i) - 1 = 3K-1 is that it?
Well is 3n1n2 - 2n1 - n2 +one a positive integer? If so, then 3(3n1n2 - 2n1 - n2 +1) - one is in am, and you've won. You lot've won the privilege of writing your very first
[tex]\mathcal{Q.E.D.}[/tex]
Bravo!
Well is 3n1n2 - 2n1 - n2 +1 a positive integer? If so, and so 3(3n1n2 - 2n1 - n2 +1) - ane is in am, and you've won. You've won the privilege of writing your very outset[tex]\mathcal{Q.E.D.}[/tex]
Bravo!
How exercise you prove it's a positive integer?
How exercise you prove it'south a positive integer?
nosotros institute that [itex](3n_1 -1)(3n_2 - 2) = 3K -1[/itex] and now we desire to show that [itex]Grand\geq ane[/itex]. Well start we notice that [itex]3n_1 -ane \geq two[/itex] and [itex]3n_2 -2\geq 1[/itex]. So [itex](3n_1 -one)(3n_2 - 2)\geq 2[/itex] Hence, [itex]3K -1 \geq 2 \Leftrightarrow 3K \geq iii \Leftrightarrow K \geq 1[/itex]
quasar987, by this way you take proved that an bm must exist an element of {an}. But, what if there exists some number ten, and some number i such that: ax i is too an chemical element of {anorthward}, merely i is not an chemical element {bm}?we constitute that [itex](3n_1 -ane)(3n_2 - 2) = 3K -ane[/itex] and at present nosotros want to show that [itex]K\geq i[/itex]. Well first nosotros notice that [itex]3n_1 -1 \geq 2[/itex] and [itex]3n_2 -2\geq 1[/itex]. Then [itex](3n_1 -1)(3n_2 - 2)\geq 2[/itex] Hence, [itex]3K -1 \geq 2 \Leftrightarrow 3K \geq iii \Leftrightarrow Grand \geq 1[/itex]
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May I suggest a little different way:
[tex]a_n \equiv two \mbox{ modern } 3, \ \forall north \in \mathbb{Northward} ^ *[/tex]
If an * bm must as well in {an}, that ways:
[tex]a_n \ b_m \equiv 2 \mbox{ modernistic } 3, \ \forall northward, grand \in \mathbb{N} ^ *[/tex]
So what'southward [tex]b_m \equiv ? \mbox{ modernistic } iii[/tex]?
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