#how i find the general term for this
fast latch
kind torrentBOT
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Let the sequence, b1, b2, b3, ... be a fibonacci-like sequence starting from 4 and 1. Wouldn't the nth term be 2^(b1)/2^(b2)*2^(b3)/2^(b4) ... 2^(bn-1)?
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Find the nth term formula for this fibonacci-like sequence and it will be done
fast latch
still don t get it
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The Fibonacci like sequence is 4, 1, 5, 6, 11, 17 ... So the nth term for the sequence is 2^(4-1+5-6+...) which stops at the (n-1)th term for term n
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still don t get it
Notice one of the first two term is 1 so the third and fourth term of this sequence are both 3 more than the original fibonnaci sequence. I think you can derive the actual sequence from there
Then calculate the sum of odd terms fibonacci sequence - sum of even terms in the fibonacci sequences to find the indices
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Proof by induction, if this sequence was true to n=k>3 and (a_(k))=2^c, a_(k- 1)=2^b, a_(k-2)=2^a, a_(k+1)=d, a_(k+2)=e then => 2a-b=e <=>c-(b-c)=e<=>c-d=e
Q.E.D
Fibonacci nth term is (1/sqrt(5))*(phi^n-(phi-sqrt(5))^n)
Our fibonacci-like sequence is (2+3), (3+3), (5+6), (8+9), (14+15)
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still don t get it
So our fibonacci sequence is (1/sqrt(5))* (phi^n-(phi-sqrt(5))^n) +(3/sqrt(5)) *(phi^(n-2)-(phi-sqrt(5))^(n-2))
[
\text{For even } n>3, \text{ nth term is: }
]
[
2^{ \left( \frac{3}{\sqrt{5}} \right) \left( \phi^{n-2}-(\phi-\sqrt{5})^{n-2} - \phi^{n-3}+(\phi-\sqrt{5})^{n-3} \right) + \left( \frac{1}{\sqrt{5}} \right) \left( \phi^n-(\phi-\sqrt{5})^n- \phi^{n-1} + (\phi-\sqrt{5})^{n-1} \right) + 7 }
]
[
\text{For odd } n>3, \text{ nth term is: }
]
[
2^{ \left( \frac{3}{\sqrt{5}} \right) \left( \phi^{n-3}-(\phi-\sqrt{5})^{n-3} - \phi^{n-2} + (\phi-\sqrt{5})^{n-2} \right) + \left( \frac{1}{\sqrt{5}} \right) \left( \phi^{n-1}-(\phi-\sqrt{5})^{n-1} - \phi^n +(\phi-\sqrt{5})^n \right) + 7 }
]
native totemBOT
lolmsnTEAL3
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The sum of all even numbers up to the term 2n in a fibonacci-like sequence is
Sum= (F_(2n+1)-F_(2n-1))+(F_(2n-1)-F_(2n-5))+...+(F_3-F_1)=F_(2n+1)-F_1
The sum of all odd numbers up to the term 2n-1 in a fibonacci-like sequence is:
Sum= (F_(2n)-F_(2n-2))+(F_(2n-2)-F_(2n-4))+...+(F_4-F_2)+F_1=F_(2n)-F_2+F1
I subtract 1 from all the exponents containing n to account for a1=1 which is not present in the sequence
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@fast latch
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I figured out the solution is related to the Fibonacci sequence then search their nth term up
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Yeah I just find out 3 more terms
so 256, 4, 8192
Write down the quotient between them
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caus b1 = 1 b2 = 16 and than b3 = 8
Then I prove it by induction
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Proof by induction, if this sequence was true to n=k>3 and (a_(k))=2^c, a_(k- 1)...
Here
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caus b1 = 1 b2 = 16 and than b3 = 8
Isn't a1=1, a2=16, a3=8? I'm using b1, b2, b3, b4 as 4, 1, 5, 6, ...
dry jetty
Yeah so that's the general strategy, try deriving it yourself