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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
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How did you come up with the formula?
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KerimF wrote: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)
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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
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Thank you, Bob.
Your steps are much better than mine.
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FelizNYC wrote:KerimF wrote: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.
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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 64However 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 = 950, 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.
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Thanks, guys.
I applied the formula and it worked, obviously.
So trial and error seems the order of the day?
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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
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