I changed the Fibonacci sequence so that the first two terms were 2. I only saw 4 powers of two in this sequence (2, 2, 4, and 16). Is there a fifth power of two in this pattern, and is there an infinite number of powers of two in this pattern?

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TopNewestI got this formula where \(f_{0}=2\) and \(f_{1}=2\): \[\dfrac{5+\sqrt{5}}{5}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-\sqrt{5}}{5}\left(\dfrac{1-\sqrt{5}}{2}\right)^n\] where \(n\) represents the \(nth\) term of the sequence. – Marc Vince Casimiro · 2 years, 3 months ago

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– Phillip Williams · 2 years, 3 months ago

Cool formula... How did you solve for that formula?Log in to reply

– Marc Vince Casimiro · 2 years, 3 months ago

Since it follows the Fibonacci recurrence relation, the general form would be \[a\left(\dfrac{1+\sqrt{5}}{2}\right)^n + b\left(\dfrac{1-\sqrt{5}}{2}\right)^n\] To get \(a\) and \(b\), you need to solve these two equations: \[a+b=c_{0}\]\[a\left(\frac{1+\sqrt{5}}{2}\right)+b\left(\frac{1-\sqrt{5}}{2}\right)=c_{1}\] where \(c_{0}\) is the first term and \(c_{1}\) is the second termLog in to reply

I have no idea. I've only reached 2692538. – Phillip Williams · 2 years, 3 months ago

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