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Another great question. +1+1+1
I recommend that you change your question to...
... This series is equal to − b a , where a and b are coprime positive integers. What is a + b ?
With an answer of 9 + 1 0 9 = 1 1 8 .
Also, is this FB group any good? I might need to get an FB account for this.
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Thank you sir.
Firstly I thought like of your recommendation however, I thought to share the same problem as it is the linked one. So I didn't make any changes.
Yes, it is a facebook group of Romanian mathematical Magazine which share problem on facebook.
I entered in the right answer but for some reason Brilliant didn't accept the negative sign. I figured I had done something incorrectly but I guess not...
Since I can't give a solution because of an error with Brilliant (even though my answer was correct), here is my solution:
The fibonacci numbers can be generated using the function F(n) = (phi1^n - phi2^n)/sqrt(5) where phi1 = (1 + sqrt(5)) / 2 and phi2 = (1 - sqrt(5)) / 2. Thus, the given summation is equivalent to (sum((-phi1^2 / 9)^n) - sum((-phi2^2 / 9)^n))/sqrt(5). Each of the sums is an infinite geometric series with r < 1 so we can use the formula 1/1-r to calculate the infinite sum. The sum therefore equals (1/(1 + phi1^2/9) + 1/(1 + phi2^2/9))/sqrt(5). I would show all the steps I took to get the final answer but it would make the comment look very messy. Through a lot of manipulation, you should get -9/109 as the final answer.
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Sorry, I cannot be helpful in case of technical issues. I hope @Brilliant Mathematics can help here.
What is ϕ − 1 ?
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reciprocal of the golden ratio. Yeah, the solution should have mentioned it.
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Thanks. Yes, the solution should have mentioned it.
I have mentioned about it. Please check it.
I know this is the cheating way but ...
I can see the sum is converging to a value -0.0825688
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You could greatly reduce the runtime if you add the following two lines at the top:
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For more information, check out this video .
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Since the generating function of Fibonacci series is F n ( x ) = n ≥ 0 ∑ F n x n = 1 − x − x 2 x , x < ϕ − 1 where ϕ is golden ratio and since it is easy to derive by it recursive relation so it is easy to note that F 2 n ( x ) n ≥ 0 ∑ F 2 n x 2 n = 2 1 ( F n ( x ) + F n ( − x ) ) giving us the last expression for F 2 n ( x ) = 2 1 ( 1 − x − x 2 x − 1 + x − x 2 x ) = x 4 − 3 x 2 + 1 x 2 now replacing x by x and them x = − 9 1 and simplification gives us n ≥ 0 ∑ ( − 1 ) n 9 n F 2 n = x 2 − 3 x + 1 x ∣ ∣ ∣ ∣ ∣ x = − 9 1 = − 1 0 9 9
We can note that, for all x < ϕ − 1 n ≥ 0 ∑ F 2 n x n = x 2 − 3 x + 1 x and n ≥ 0 ∑ F 2 n + 1 x n = x 2 − 3 x + 1 1 − x