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Still i am not sure how to solve it. Is it some kind of arithmetic sequence or geometric sequence?.Scott said:@engnrshyckh:
I will give you these clues:
1) The pyramid shape is irrelevant. The sequence 2, 3, 6, 8, 15, 22, 38, x, 98, 156 is sufficient.
2) If the terms are labelled ##A_0=2##, ##A_1=3##, ##A_2=6##, etc; then the even A's are computed with a slightly different formula than the odd A's.
3) ##A_{-1}## is not zero.
While i did use this initially to find the solution, you can also give a single equation valid for both even and odd A's. (and no [itex] (-1)^n [/itex] or anything like that in it).Scott said:2) If the terms are labelled A0=2, A1=3, A2=6, etc; then the even A's are computed with a slightly different formula than the odd A's.
Using ##(-1)^n## obviously works. I also think there is a way to use absolute value. But for the purpose of providing a clue, those aren't very useful.To @engnrshyckh:willem2 said:While i did use this initially to find the solution, you can also give a single equation valid for both even and odd A's. (and no [itex] (-1)^n [/itex] or anything like that in it)
I said i was not using that..Scott said:Using ##(-1)^n## obviously works. I also think there is a way to use absolute value. But for the purpose of providing a clue, those aren't very useful.
There is an exact non-recursive formula for the elements of the Fibonacci series, which does use the exponential function..Scott said:However, note that, although the Fibonacci series is exponential, no exponential function is used to compute its elements.
I see. You go back one more term to get the even/odd information.willem2 said:If you type [insert recurrence relation here], a(0)=2, a(1)=3, a(2) = 6 in wolfram alpha, you'll get the exact solution.
Clever. I missed thatmfb said:You can calculate the differences between two subsequent numbers and check if these are somehow linked to other elements in the sequence.
These are successive ratio 1.5,2,1.333,1. 875 but these are not the same as in geometric propagation.willem2 said:While i did use this initially to find the solution, you can also give a single equation valid for both even and odd A's. (and no [itex] (-1)^n [/itex] or anything like that in it)
Why are you calculating ratios?engnrshyckh said:These are successive ratio 1.5,2,1.333,1. 875 but these are not the same as in geometric propagation.
mfb said:You can calculate the differences between two subsequent numbers and check if these are somehow linked to other elements in the sequence.
Difference is not the same also 1,3,2,7,7... To use arithmetic propagation farmulaDrClaude said:Why are you calculating ratios?
engnrshyckh said:Difference is not the same also 1,3,2,7,7... To use arithmetic propagation farmula
118 is the correct ans thanks for the helpPeroK said:Here's a clue: it's not a Fibonacci sequence!
If i am not wrongengnrshyckh said:118 is the correct ans thanks for the help
It can't be ##118##. That's too big.engnrshyckh said:118 is the correct ans thanks for the help
OK. But I gave you several more numbers in the sequence. Calculate all of the ratios for pairs up to 19307/11934. What you will discover is that it IS an exponential sequence - very similar to Fibonacci.engnrshyckh said:These are successive ratio 1.5,2,1.333,1. 875 but these are not the same as in geometric propagation.
Then what i am missing?PeroK said:It can't be ##118##. That's too big.
He's calculating ratios because it is a possible method for detecting an exponential sequence (suggested by me). The only proble is that he didn't calculate enough of them. The will close on the Golden Ratio - a significant clue to how the sequence can be generated.DrClaude said:Why are you calculating ratios?
How does 118 work out? I don't see the pattern.engnrshyckh said:Then what i am missing?
Yes its an exponential sequence but i am unable to find its equation which gives the correct answer.Scott said:OK. But I gave you several more numbers in the sequence. Calculate all of the ratios for pairs up to 19307/11934. What you will discover is that it IS an exponential sequence - very similar to Fibonacci.
I have something a lot simpler! Are there two answers?.Scott said:He's calculating ratios because it is a possible method for detecting an exponential sequence (suggested by me). The only proble is that he didn't calculate enough of them. The will close on the Golden Ratio - a significant clue to how the sequence can be generated.
It's not just an exponential equation. The ratio is the about the same as for the Fibonacci Series. So you should ask - how does the formula for generating the next Fibonacci number (##v_i = v_{i-1} + v_{i-2}##) result in the Golden Ratio?engnrshyckh said:Yes its an exponential sequence but i am unable to find its equation which gives the correct answer
Enlighten me i shall be greatfull.Scott said:It's not just an exponential equation. The ratio is the about the same as for the Fibonacci Series. So you should ask - how does the formula for generating the next Fibonacci number (##v_i = v_{i-1} + v_{i-2}##) result in the Golden Ratio?
I stopped typing my clue in as soon as I saw yours pop up.PeroK said:One last clue:
## 2 + 3 \approx 6##
##3 + 6 \approx 8##
##6 + 8 \approx 15##
Thanks got it in some terms we add and aubtract 1 alternatively.Scott said:We're not suppose to just give you the solution. So here's another (hopefully final) clue:
We know (2, 3, 6, 8, 15, ...) isn't Fibonacci because:
6 <> 2+3 (just misses)
8 <> 3+6 (just misses)
15 <> 6+8 (just misses)
22 <>
And now that a solution is finally revealed, @willem2 solved it with this:willem2 said:You can however get an exact formula for the n-th element of the series, and it does involve (-1)^n.
If you type [insert recurrence relation here], a(0)=2, a(1)=3, a(2) = 6 in wolfram alpha, you'll get the exact solution.
It is also easy to show that both approaches are the same. Take.Scott said:And now that a solution is finally revealed, @willem2 solved it with this:
## v_i = 2v_{i-2}+v_{i-3} ##