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paulb203

Member

Registered: 2023-02-24

Posts: 136

Sequences; finding the formula that gives us the nth term

I've been given this question.

Here is a sequence;

2,5,11,23,…

Find the next two terms.

*

I can see that each term is 2 times the previous term, plus 1.
But I can’t find the formula for the nth term.

I know that with arithmetic sequences the difference is the same each time
I know that with a geometric sequence the terms double, each time, or treble each time, or, etc, etc, and the formula is Un=ar^n-1
I know that with a quadratic sequence the second difference is the same each time and the formula is an^2+bn+c

But I don't know how to find the formula for the nth term for this sequence

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

Re: Sequences; finding the formula that gives us the nth term

nycguitarguy

Member

Registered: 2024-02-24

Posts: 493

Re: Sequences; finding the formula that gives us the nth term

How did you come up with the formula?

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

Re: Sequences; finding the formula that gives us the nth term

How did you come up with the formula?

I did it by following primitive steps since I forgot, at age 75, the advanced ones.

f(n)   = f(n-1)*2+1				
f(n-1) = f(n-2)*2+1				
	f(n)   = (f(n-2)*2+1)*2+1			
	f(n-2) = f(n-3)*2+1			
		f(n)   = ((f(n-3)*2+1)*2+1)*2+1		
		f(n-3) = f(n-4)*2+1		
			f(n)   = (((f(n-4)*2+1)*2+1)*2+1)*2+1	
			f(n-4) = f(n-5)*2+1	
				f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1
				
f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2+1)*2*2+2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2*2*2+2*2+2+1				
f(n) = (((f(n-5)*2+1)*2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-5)*2*2*2*2*2+2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-a)*2^a… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1				

Um=k*r^(m-1)				
In the following series
… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1
we have
k=1
m=a			
f(n) = f(n-a)*2^a+2^(a)-1				
f(n) = 2^a*[f(n-a)+1]-1				

Let us assume:
n-a=1				
n=a+1
f(a+1) = 2^a*[f(1)+1]-1				
But
f(1)= 2				
Therefore
f(a+1) = 2^a*3-1				

Again, let us assume:
a+1=n				
a=n-1				
f(n) = 2^(n-1)*3-1				
f(n) = 3*2^(n-1)-1				

Last edited by KerimF (2024-03-18 00:42:28)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,172

Re: Sequences; finding the formula that gives us the nth term

I got it like this:

If the numbers are doubling then the formula ought to involve powers of two

      n          power of 2

      1                2
      2                4
      3                8
      4               16
      5               32
      6               64

However this fails to generate the right sequence.  Each term needs a further amount so I wrote down this amount.

2   +  0  = 2
4   +  1  = 5
8   +  3  = 11
16 +  7  = 23
32 + 15 = 47
64 + 31 = 95

0, 1, 3 etc are one short of the previous power of 2, so I ended up with

This is the same as KerimF's as the following shows:

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

Re: Sequences; finding the formula that gives us the nth term

Thank you, Bob.
Your steps are much better than mine.

nycguitarguy

Member

Registered: 2024-02-24

Posts: 493

Re: Sequences; finding the formula that gives us the nth term

How did you come up with the formula?

I did it by following primitive steps since I forgot, at age 75, the advanced ones.

f(n)   = f(n-1)*2+1				
f(n-1) = f(n-2)*2+1				
	f(n)   = (f(n-2)*2+1)*2+1			
	f(n-2) = f(n-3)*2+1			
		f(n)   = ((f(n-3)*2+1)*2+1)*2+1		
		f(n-3) = f(n-4)*2+1		
			f(n)   = (((f(n-4)*2+1)*2+1)*2+1)*2+1	
			f(n-4) = f(n-5)*2+1	
				f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1
				
f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2+1)*2*2+2+1				
f(n) = (((f(n-5)*2+1)*2+1)*2*2*2+2*2+2+1				
f(n) = (((f(n-5)*2+1)*2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-5)*2*2*2*2*2+2*2*2*2+2*2*2+2*2+2+1				
f(n) = f(n-a)*2^a… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1				

Um=k*r^(m-1)				
In the following series
… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1
we have
k=1
m=a			
f(n) = f(n-a)*2^a+2^(a)-1				
f(n) = 2^a*[f(n-a)+1]-1				

Let us assume:
n-a=1				
n=a+1
f(a+1) = 2^a*[f(1)+1]-1				
But
f(1)= 2				
Therefore
f(a+1) = 2^a*3-1				

Again, let us assume:
a+1=n				
a=n-1				
f(n) = 2^(n-1)*3-1				
f(n) = 3*2^(n-1)-1				

Your steps are too detailed. There gotta be an easier way.

nycguitarguy

Member

Registered: 2024-02-24

Posts: 493

Re: Sequences; finding the formula that gives us the nth term

I got it like this:

If the numbers are doubling then the formula ought to involve powers of two

      n          power of 2

      1                2
      2                4
      3                8
      4               16
      5               32
      6               64

However this fails to generate the right sequence.  Each term needs a further amount so I wrote down this amount.

2   +  0  = 2
4   +  1  = 5
8   +  3  = 11
16 +  7  = 23
32 + 15 = 47
64 + 31 = 95

0, 1, 3 etc are one short of the previous power of 2, so I ended up with

This is the same as KerimF's as the following shows:

Bob

Your steps are so much better.

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Sequences; finding the formula that gives us the nth term

Thanks, guys.
I applied the formula and it worked, obviously.
So trial and error seems the order of the day?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,172

Re: Sequences; finding the formula that gives us the nth term

I prefer to call it trial and improvement.

I do most integration questions by trying a likely answer and differentiating to see if I'm right.  And, if you can spot an answer just by looking, then show it fits, then that's an acceptable way to answer a question.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob