C1 Sequences help please :/ Watch

Super199
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Can someone explain how to work these out thanks.
Suggest possible recurrence relationships for the following sequences :

A). 3,5,7,9 (adding 2 each time)
B). 100,25,6.25,1.5625 (dividing by 4)
C) 3,7,15,31 ( adding on double the difference. so 4,8,16,32,64 etc..)
j). 4,10,18,38,74 ( not sure)

I can work out how the pattern changes but don't know how to put it in the correct form. I.E un= ...
Any help would be appreciated as I missed some of the lesson. Thanks
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the bear
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the last one involves doubling and then either adding or subtracting a certain amount
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Super199
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(Original post by the bear)
the last one involves doubling and then either adding or subtracting a certain amount
Is it (x2+2),(x2-2),(x2+2)(x2-2) etc.? How do I do a recurrence relationship?
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the bear
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un+1 = 2*un + 2*(?n)

you need to find something suitable for ?
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Super199
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(Original post by the bear)
un+1 = 2*un + 2*(?n)

you need to find something suitable for ?
I dunno
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the bear
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(Original post by Super199)
I dunno
i am not allowed to tell you :eek:
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Super199
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(Original post by the bear)
i am not allowed to tell you :eek:
How do I get the answer then? Talk me through it if you can? I guess that isn't giving the answer away.
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zeldasmockingjay
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I'm literally doing this chapter atm, so I thought I'd attempt to lend a hand... but I only know the first one...

A) Un=2n+1

Other than that, I have no clue!! But, I hope that that little bit helps at least!
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517340
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For A, the sequence is 2n+1. You've correctly identified the pattern, which is basically the two times table, plus one. Hence 2n+1

For B, the sequence is 400/4n. When sequences multiply, the n would be as a power of the amount it is multiplied by. To get the 400, you go one step back. This goes for all sequences, for finding the initial value.
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Super199
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I sort of understand but the book says the answer for 1 is. uk+1= Uk+ 2, U1=3
Anyone care to explain how you get it in this form?
I understand the formula but I don't get how to put sequences into this.
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517340
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C: Let's look at the second sequence, it's basically 2n, but forwards one term (it should start on 1), so that makes it 2n+1. How do I know it's 2n? You multiply that second sequence by 2 each time (Remember what I said about multiplying? 'n' becomes a power of that number). Now, compare that to the first.

SEQ1: 3 - 7 - 15 - 31
SEQ2: 4 - 8 - 16 - 32

It's the same sequence, but -1. So your formula is 2n+1-1

Last one's got me stumped for the moment!
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Super199
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(Original post by Ruthless Dutchman)
C: Let's look at the second sequence, it's basically 2n, but forwards one term (it should start on 1), so that makes it 2n+1. How do I know it's 2n? You multiply that second sequence by 2 each time (Remember what I said about multiplying? 'n' becomes a power of that number). Now, compare that to the first.

SEQ1: 3 - 7 - 15 - 31
SEQ2: 4 - 8 - 16 - 32

It's the same sequence, but -1. So your formula is 2n+1-1

Last one's got me stumped for the moment!
Ah great thanks! I sort of understand now. Have you got an answer to the last one?
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517340
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(Original post by Super199)
I sort of understand but the book says the answer for 1 is. uk+1= Uk+ 2, U1=3
Anyone care to explain how you get it in this form?
I understand the formula but I don't get how to put sequences into this.
I've never worked with that form before!

But from what i can see, it's saying that your next value uk+1 is your current value uk plus 2, where the first value u1 is 3
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TenOfThem
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(Original post by zeldasmockingjay)
I'm literally doing this chapter atm, so I thought I'd attempt to lend a hand... but I only know the first one...

A) Un=2n+1

Other than that, I have no clue!! But, I hope that that little bit helps at least!

(Original post by Ruthless Dutchman)
C: Let's look at the second sequence, it's basically 2n, but forwards one term (it should start on 1), so that makes it 2n+1. How do I know it's 2n? You multiply that second sequence by 2 each time (Remember what I said about multiplying? 'n' becomes a power of that number). Now, compare that to the first.

SEQ1: 3 - 7 - 15 - 31
SEQ2: 4 - 8 - 16 - 32

It's the same sequence, but -1. So your formula is 2n+1-1

Last one's got me stumped for the moment!
The OP need recurrence relationships NOT general terms
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TenOfThem
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(Original post by Super199)
Can someone explain how to work these out thanks.
Suggest possible recurrence relationships for the following sequences :

A). 3,5,7,9 (adding 2 each time)
B). 100,25,6.25,1.5625 (dividing by 4)
C) 3,7,15,31 ( adding on double the difference. so 4,8,16,32,64 etc..)
j). 4,10,18,38,74 ( not sure)

I can work out how the pattern changes but don't know how to put it in the correct form. I.E un= ...
Any help would be appreciated as I missed some of the lesson. Thanks
Do you understand what the notation

U_1. U_k, U_{k+1}

mean
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517340
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(Original post by Super199)
Ah great thanks! I sort of understand now. Have you got an answer to the last one?
That last one was fiendish.

Looking at the sequence I struggled to see any connections, even at the second level

SEQ1: 4 --- 10 --- 18 --- 38 --- 74
SEQ2:-- +6 -- +8 -- +20 -- +36

At least until I noticed that the link from the third to the fourth term in SEQ1 is double the second term.

Leading me to the conclusion that the next term un+1 is the current term un plus double the previous term 2un-1

To get from 4 to 10, it would be 10 = 4 + 2*? where ? would be the previous term.
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Super199
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(Original post by TenOfThem)
Do you understand what the notation

U_1. U_k, U_{k+1}

mean
U1 means first number of sequence?
Uk is the term I think
Uk+1 is the term after so U2?
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brianeverit
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(Original post by Super199)
Can someone explain how to work these out thanks.
Suggest possible recurrence relationships for the following sequences :

A). 3,5,7,9 (adding 2 each time)
B). 100,25,6.25,1.5625 (dividing by 4)
C) 3,7,15,31 ( adding on double the difference. so 4,8,16,32,64 etc..)
j). 4,10,18,38,74 ( not sure)

I can work out how the pattern changes but don't know how to put it in the correct form. I.E un= ...
Any help would be appreciated as I missed some of the lesson. Thanks
For the last one you are multiplying by 2 and either adding 2 or subtracting 2, which you achieve like this (-1)^n2\mathrm{\ adds\ } 2 \mathrm{\ if\ } n \mathrm{\ is\ even\ but\  subtracts\ } 2 \mathrm{\ if\ } n \mathrm{\ is\ odd\ }.
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