k = 0 ∑ ∞ F 2 k 1
Evaluate the sum above, where F n is the n th Fibonacci number . Round your answer to the second decimal place.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Nice. I am not well versed with the ∑ stuff, but it looks cool.
This converges extremely quickly. F 2 6 > 1 0 1 3 so the first 7 terms give a sum 2 . 3 8 1 9 6 6 0 1 1 on my calculator which is accurate to all displayed digits.
Out of interest, I put this into Inverse Symbolic Calculator which had the single hit 2 7 − 5 which seems reasonable as the exact solution.
Problem Loading...
Note Loading...
Set Loading...
Using Binet's Formula for the Fibonacci numbers, F n = 5 1 [ ϕ n − ( − ϕ − 1 ) n ] n ≥ 0 where ϕ = 2 1 ( 1 + 5 ) is the golden ratio, we see that S = n = 0 ∑ ∞ F 2 n = n = 0 ∑ ∞ ϕ 2 n − ( − ϕ − 1 ) 2 n 5 = n = 0 ∑ ∞ ϕ 2 n 5 ( 1 − ( − ϕ − 2 ) 2 n ) − 1 = ϕ ( 1 + ϕ − 2 ) 5 + n = 1 ∑ ∞ ϕ 2 n 5 ( 1 − ϕ − 2 n + 1 ) − 1 = ϕ ( 1 + ϕ − 2 ) 5 + n = 1 ∑ ∞ m = 0 ∑ ∞ ϕ ( 2 m + 1 ) 2 n 5 Now the numbers ( 2 m + 1 ) 2 n for m ≥ 0 and n ≥ 1 simply run through all the even positive integers. Thus we deduce that S = ϕ ( 1 + ϕ − 2 ) 5 + n = 1 ∑ ∞ ϕ 2 n 5 = ϕ ( 1 + ϕ − 2 ) 5 + 1 − ϕ − 2 5 ϕ − 2 = 1 + ϕ 2 5 ϕ + ϕ 2 − 1 5 = ϕ + 2 5 ( 3 ϕ − 1 ) = 2 1 ( 7 − 5 )
@Jeremy Galvagni