C1 Sequences help please :/

the last one involves doubling and then either adding or subtracting a certain amount

Reply 2

OP

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?

Reply 3

the bear

u_n+1 = 2*u_n + 2*(?ⁿ)

you need to find something suitable for ?

Reply 4

OP

Original post by the bear

u_n+1 = 2*u_n + 2*(?ⁿ)

you need to find something suitable for ?

I dunno

Reply 5

the bear

Original post by Super199

I dunno

i am not allowed to tell you :eek:

Reply 6

OP

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.

Reply 7

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!

Reply 8

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.

Reply 9

OP

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.

(edited 10 years ago)

Reply 10

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!

Reply 11

OP

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?

Reply 12

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 u_k+1 is your current value u_k plus 2, where the first value u₁ is 3

Reply 13

TenOfThem

18

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

Reply 14

TenOfThem

18

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

Reply 15

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 u_n+1 is the current term u_n plus double the previous term 2u_n-1

To get from 4 to 10, it would be 10 = 4 + 2*? where ? would be the previous term.

Reply 16

OP