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The Fibonacci Numbers are defined as F1 = F2 = 1 and Fn+1 = Fn + Fn-1. Calculate the following infinite sum: F1 3F2 32F3 3nFn+1 --- + ---- + ----- + ... + ----- + ... 1 5 52 5n |
| Answer: 25
Solution: Let A = F1 + (3/5)F2 + (32/52)F3 + ... A = F1 + (3/5)F2 + (32/52)(F1+F2) + (33/53)(F2+F3) + ... A = 1 + 3/5 + (32/52)F1 + (32/52)F2 + (33/53)F2 + (33/53)F3 + ... A = 1 + (3/5)F1 + (32/52)F1 + (32/52)F2 + (33/53)F2 + (33/53)F3 + ... Split A into two pieces, taking alternate terms for A1 and A2,
You can see that
So A = 25 |
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There's nothing special about the ratio 3/5, except that it's less than 1/φ, where φ (phi) is the golden ratio, (sqrt(5)+1)/2. The sum
∞
∑
k=0rk Fk converges iff |r|<1/φ,
because the ratio of successive Fibonacci numbers approaches φ. If you revisit the proof, above, replacing each instance of 3/5 by r, then you will soon arrive at the general solution
∞
∑
k=0rk Fk = 1/(1-r-r²).
The webmaster and author of this Math Help site is Graeme McRae.