Math Is Fun Forum

paulb203

Member

Registered: 2023-02-24

Posts: 137

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: 168

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

nycguitarguy

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 566

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

How did you come up with the formula?

---

The Rapture is the central event in biblical prophecy. The greatest truth about the Rapture is not its timing but it's reality. The Rapture will be the great disappearance.

Dr. David Jeremiah

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 168

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: 10,208

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: 168

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

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

nycguitarguy

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 566

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.

---

The Rapture is the central event in biblical prophecy. The greatest truth about the Rapture is not its timing but it's reality. The Rapture will be the great disappearance.

Dr. David Jeremiah

nycguitarguy

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 566

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.

---

The Rapture is the central event in biblical prophecy. The greatest truth about the Rapture is not its timing but it's reality. The Rapture will be the great disappearance.

Dr. David Jeremiah

paulb203

Member

Registered: 2023-02-24

Posts: 137

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: 10,208

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