Infinite Sum of Series Which Switches Signs and ONLY Occurs on Fibonacci Numbers

NeeleshGupta023

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Hey people,
The beginning of a series is shown in a problem.

(7)/(1*1-1) - (7)/(2*2-1) + (7)/(3*3-1) - (7)/(5*5-1) + (7)/(8*8-1) - ...

I had tried to input this into a sigma sum, but I do not know how to only get it to occur on Fibonacci numbers.
This is what I have so far

((7)(-(-1)^k)) / ((k-1)(k+1))
 
Upon further research, I found the equation for Fibonacci number to be ((Phi^n) - ((-1/Phi)^n))/(sqrt(5)).
I used this to deduce this sigma sum formula. Would this work?
 

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Hey people,
The beginning of a series is shown in a problem.

(7)/(1*1-1) - (7)/(2*2-1) + (7)/(3*3-1) - (7)/(5*5-1) + (7)/(8*8-1) - ...

I had tried to input this into a sigma sum, but I do not know how to only get it to occur on Fibonacci numbers.
This is what I have so far

((7)(-(-1)^k)) / ((k-1)(k+1))
Are you sure that is correct? That denotes division by 0 - an illegal operation!
 
No, that is not correct.
After further inspection of the problem, I have found that the -1 alternates from a plus one to minus one
This is the problem
 

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After much thinking, I have came up with this as the equation for the sum, but I do not know how to solve it.

[MATH]\sum _{n=2}^{\infty }\:\left(\frac{7\left(-1\right)^n}{\left(\frac{\left(\phi ^n-\left(\frac{-1}{\phi \:\:\:}\right)^n\right)}{\sqrt{5}}\right)^2+\left(-1\right)^n}\right)[/MATH]
 
What you did looks fine. What are you trying to solve? Initially you said that you wanted to write the series in sigma notation which you did.
You really need to tell us the complete problem all at once instead of piece by piece as we go along.
 
I figured that but we still like for the student to ask their questions. I do not think that your question has a simple answer. Possibly others might have a nice method.
Ahahahaha!! I FIGURED IT OUT! I was able to input the sum into my TI-84, and instead of finding the sum from 2 to 10^99 or an actual infinity, I did it from 2 to 100, and I was able to find the correct answer, and I did not even use the Phi button, and I had rounded Phi to 8 decimal points.
 
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