# Finding the simplest equation for the sequence: 1, 5, 21, 85... (Challenge for pros)

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#1
Obviously, the number is going up by 4^(n-1) each time.

1st term) Let the first term = 1
2nd term) The term before + 4^1 = 5
3rd term) The term before + 4^2 = 21
4th term) The term before + 4^3 + 85

And so forth.

Without using trial and error, what is the easiest and best method to find a simplified equation for similar questions (e.g. sequences with differences going up by powers)?

I don't want the equation to rely on other terms, e.g. I don't want to include the "The term before +" bit. I want it to be a formula, so you can calculate the 800th term without calculating any other terms.

I am not doing this for education but to expand my maths, I just do not see a method how. I know the answer but not the method. I just don't like doing it via trial and error.

Thanks!

EDIT: This is the answer. The challenge is to find the EASIEST and most reliable method of finding the equation of a sequence with differences of increasing powers, for instance, the one above is increasing by powers of 4. I have yet to find the easiest method, I have tried Google and messing about, yet my best method is simply trial and error and including (4^something) in the equation; however, I am certain there is a better method.
1
7 years ago
#2
(Original post by Konnichiwa)
Obviously, the number is going up by 4^(n-1) each time.

1st term) Let the first term = 1
2nd term) The term before + 4^1 = 5
3rd term) The term before + 4^2 = 21
4th term) The term before + 4^3 + 85

And so forth.

Without using trial and error, what is the easiest and best method to find a simplified equation for similar questions (e.g. sequences with differences going up by powers)?

I don't want the equation to rely on other terms, e.g. I don't want to include the "The term before +" bit. I want it to be a formula, so you can calculate the 800th term without calculating any other terms.

I am not doing this for education but to expand my maths, I just do not see a method how. I know the answer but not the method. I just don't like doing it via trial and error.

Thanks!
Odd choice of rule I would have gone with Still a term-to-term rule though
0
7 years ago
#3
Just a question, what year are you in?
0
7 years ago
#4
(Original post by Konnichiwa)
Obviously, the number is going up by 4^(n-1) each time.

1st term) Let the first term = 1
2nd term) The term before + 4^1 = 5
3rd term) The term before + 4^2 = 21
4th term) The term before + 4^3 + 85

And so forth.

Without using trial and error, what is the easiest and best method to find a simplified equation for similar questions (e.g. sequences with differences going up by powers)?

I don't want the equation to rely on other terms, e.g. I don't want to include the "The term before +" bit. I want it to be a formula, so you can calculate the 800th term without calculating any other terms.

I am not doing this for education but to expand my maths, I just do not see a method how. I know the answer but not the method. I just don't like doing it via trial and error.

Thanks!
Like in this case ?
0
#5
Yes, but it can be expressed without relying on other terms.

There is an answer, and I am wondering if there is a method that people use to find expressions (without relying on other terms) of a sequence with differences going up in powers/exponentially.

This is what troubled me.
0
7 years ago
#6
Duuuuuuurrrrrrrrr me

Good spot
0
#7
Yes, that is the answer but I cannot find the method.

(Original post by #Unknown)
Just a question, what year are you in?
13.
0
7 years ago
#8
(Original post by TenOfThem)
Duuuuuuurrrrrrrrr me

Good spot
I'd like to know the answer to this tbh. I'm not the best of coming up with equations for series. DO you have a method?
0
7 years ago
#9
(Original post by Konnichiwa)
Yes, but it can be expressed without relying on other terms.

yes, I realise that - hence the second part of my post
0
7 years ago
#10
(Original post by Konnichiwa)
Yes, that is the answer but I cannot find the method.

13.
It's always the common difference, minus the first number, divided by some ratio. The common difference in this series is .
It's not always minus the first term. I don't really have a set method. Just practice it a bit.
1
#11
(Original post by keromedic)
I'd like to know the answer to this tbh. I'm not the best of coming up with equations for series. DO you have a method?
No, I don't have the method, but I feel that such a method exists.

It should be a nice challenge finding the method.
0
7 years ago
#12
(Original post by keromedic)
I'd like to know the answer to this tbh. I'm not the best of coming up with equations for series. DO you have a method?
In this question - easy

I should have left well alone with the OPs original formula

Then think about summing as it is just a GP being summed
0
#13
(Original post by keromedic)
It's always the common difference, minus the first number, divided by some ratio. The common difference in this series is Thanks, I'll test that method soon!
0
7 years ago
#14
(Original post by TenOfThem)
In this question - easy

I should have left well alone with the OPs original formula

Then think about summing as it is just a GP being summed
Fair enough.
0
7 years ago
#15
(Original post by majmuh24)
That's the way you are supposed to do it, but instead of the term before, you use u(n) [nth term]=u(n-1)[the term before] + 4^(n-1), with u(1)[first term] being 1, does this clear things up for you?

BTW defining a sequence like this is called a RECURRENCE RELATION I think:-)

Posted from TSR Mobile
0
7 years ago
#16
(Original post by Konnichiwa)
Yes, that is the answer but I cannot find the method.

13.
It can also be written as a cubic equation. 0
7 years ago
#17
Interesting

That is a different sequence to the one the rest of us have found

But, of course, with only a finite number of terms there will be different rules
0
7 years ago
#18

How do you come up with them? Seems like a very impressive skill IMO! 0
7 years ago
#19
OP, perhaps give us another sequence?
0
7 years ago
#20
(Original post by krisshP)
How do you come up with them? Seems like a very impressive skill IMO! I should have pointed out that the cubic only holds for the 4 values given. It just shows that you cannot uniquely define a sequence by specifying a few values. You have to ngiove some kind of formula or recurrence relation.
0
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