Fibonacci Fraction

Algebra Level 5

∑ i = 1 4302 ( 3 + ( − 1 ) i ) F i ∑ i = 1 2151 F 2 i 2 + ∑ i = 1 1434 F 3 i = ? \dfrac{\displaystyle\sum_{i=1}^{4302}(3+(-1)^{i})F_{i}}{\displaystyle\sum_{i=1}^{2151}\frac{F_{2i}}{2}+\displaystyle\sum_{i=1}^{1434}F_{3i}} =\ ?

Notation: F n F_n denotes the n n th Fibonacci number that is defined as F 1 = 1 F_1=1 , F 2 = 1 F_2 = 1 , and F n = F n − 1 + F n − 2 F_n = F_{n-1} + F_{n-2} for n ≥ 3 n \ge 3 .


The answer is 4.

Braden Dean by Braden Dean , 18, USA

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3 solutions

Chew-Seong Cheong
Nov 30, 2020

We can easily prove by induction that

∑ i = 1 n F 2 i − 1 = F 2 n , ∑ i = 1 n F 2 i = F 2 n + 1 − 1 , and ∑ i = 1 n F 3 i = F 3 n + 2 − 1 2 \sum_{i=1}^n F_{2i-1} = F_{2n}, \quad \sum_{i=1}^n F_{2i} = F_{2n+1} - 1, \text{ and } \sum_{i=1}^n F_{3i} = \frac {F_{3n+2} - 1}2

Let n = 717 n=717 ; then 2 n = 1434 2n = 1434 , 3 n = 2151 3n = 2151 , and 6 n = 4302 6n=4302 . And

∑ i = 1 6 n ( 3 + ( − 1 ) i ) F i ∑ i = 1 3 n 1 2 F 2 i + ∑ i = 1 2 n F 3 i = 2 ∑ i = 1 3 n F 2 i − 1 + 4 ∑ i = 1 3 n F 2 i 1 2 ∑ i = 1 3 n F 2 i + ∑ i = 1 2 n F 3 i = 2 F 6 n + 4 F 6 n + 1 − 4 1 2 F 6 n + 1 − 1 2 + 1 2 F 6 n + 2 − 1 2 = 4 F 6 n + 8 F 6 n + 1 − 8 F 6 n + 1 + F 6 n + 2 − 2 = 4 F 6 n + 4 F 6 n + 1 + 4 F 6 n + 1 − 8 F 6 n + 1 + F 6 n + 2 − 2 = 4 F 6 n + 1 + 4 F 6 n + 2 − 8 F 6 n + 1 + F 6 n + 2 − 2 = 4 \begin{aligned} \frac {\sum_{i=1}^{6n}(3+(-1)^i)F_i}{\sum_{i=1}^{3n}\frac 12 F_{2i} + \sum_{i=1}^{2n}F_{3i}} & = \frac {2 \sum_{i=1}^{3n}F_{2i-1} + 4 \sum_{i=1}^{3n}F_{2i}}{\frac 12 \sum_{i=1}^{3n} F_{2i} + \sum_{i=1}^{2n}F_{3i}} \\ & = \frac {2F_{6n}+4F_{6n+1}-4}{\frac 12F_{6n+1}-\frac 12 + \frac 12 F_{6n+2}-\frac 12} \\ & = \frac {4F_{6n}+8F_{6n+1}-8}{F_{6n+1} + F_{6n+2}-2} \\ & = \frac {\blue{4F_{6n}+4F_{6n+1}} +4F_{6n+1}-8}{F_{6n+1} + F_{6n+2}-2} \\ & = \frac {4F_{6n+1}+\blue{4F_{6n+2}}-8}{F_{6n+1} + F_{6n+2}-2} \\ & = \boxed 4 \end{aligned}

0 Helpful 1 Interesting 5 Brilliant 1 Confused
Lingga Musroji
Dec 1, 2020

Let ∑ i = 1 2151 F 2 i = x \sum_{i=1}^{2151}F_{2i}=x and ∑ i = 1 2151 F 2 i − 1 = y \sum_{i=1}^{2151}F_{2i-1}=y

Numerator:

∑ i = 1 4302 ( 3 + ( − 1 ) i ) F i = ∑ i = 1 2151 4 F 2 i + ∑ i = 1 2151 2 F 2 i − 1 = 4 x + 2 y \sum_{i=1}^{4302}(3+(-1)^i)F_i=\sum_{i=1}^{2151}4F_{2i}+\sum_{i=1}^{2151}2F_{2i-1}=4x+2y

Denominator:

∑ i = 1 1434 F 3 i = F 3 + F 6 + F 9 + … + F 4302 ∑ i = 1 1434 F 3 i = F 1 + F 2 + F 4 + F 5 + F 7 + F 9 + … + F 4300 + F 4301 ∑ i = 1 1434 F 3 i = ( F 1 + F 3 + F 5 + F 7 + F 9 + … + F 4301 ) + ( F 2 + F 4 + F 6 + F 8 + … + F 4302 ) − ( F 3 + F 6 + F 9 + … + F 4302 ) ∑ i = 1 1434 F 3 i = x + y − ∑ i = 1 1434 F 3 i 2 ∑ i = 1 1434 F 3 i = x + y ∑ i = 1 1434 F 3 i = x + y 2 \begin{array}{ccc}\sum_{i=1}^{1434}F_{3i}&=&F_3+F_6+F_9+\ldots+F_{4302}\\\sum_{i=1}^{1434}F_{3i}&=&F_1+F_2+F_4+F_5+F_7+F_9+\ldots+F_{4300}+F_{4301}\\\sum_{i=1}^{1434}F_{3i}&=&(F_1+F_3+F_5+F_7+F_9+\ldots+F_{4301})+(F_2+F_4+F_6+F_8+\ldots+F_{4302})-(F_3+F_6+F_9+\ldots+F_{4302})\\\sum_{i=1}^{1434}F_{3i}&=&x+y-\sum_{i=1}^{1434}F_{3i}\\2\sum_{i=1}^{1434}F_{3i}&=&x+y\\\sum_{i=1}^{1434}F_{3i}&=&\frac{x+y}{2}\end{array}

∑ i = 1 2151 F 2 i 2 + ∑ i = 1 1434 F 3 i = x 2 + x + y 2 = 2 x + y 2 \sum_{i=1}^{2151}\frac{F_{2i}}{2}+\sum_{i=1}^{1434}F_{3i}=\frac{x}{2}+\frac{x+y}{2}=\frac{2x+y}{2}

Final answer:

4 x + 2 y 2 x + y 2 = 4 \frac{4x+2y}{\frac{2x+y}{2}}=4

0 Helpful 0 Interesting 1 Brilliant 1 Confused
Braden Dean
Nov 29, 2020

∑ i = 1 4302 ( ( 3 + ( − 1 ) i ) F i ) \displaystyle\sum_{i=1}^{4302}\Big((3+(-1)^{i})F_{i}\Big)

can be rewritten as:

∑ i = 1 4302 2 F i + 2 ∑ i = 1 4302 ( 1 + ( − 1 ) i ) F i 2 \displaystyle\sum_{i=1}^{4302}2F_{i} + 2\displaystyle\sum_{i=1}^{4302}\frac{(1+(-1)^{i})F_{i}}{2}

This can then be evaluated to

2 F 4304 − 2 + 2 F 4303 − 2 = 2 F 4305 − 4 2F_{4304} - 2 + 2F_{4303} - 2 = 2F_{4305} -4

as the sum of the first n Fibonacci numbers is equal to F n + 2 − 1 F_{n+2}-1 , and the sum of every other Fibonacci number starting at F 2 F_2 is equal to F n + 1 − 1 F_{n+1} - 1

We can then evaluate the bottom, which is much simpler:

∑ i = 1 2151 F 2 i 2 = F 4303 − 1 2 \displaystyle\sum_{i=1}^{2151}\frac{F_{2i}}{2} = \frac{F_{4303}-1}{2}

Every third Fibonacci can be evaluated as

S = F 3 + F 6 + F 9 + . . . + F n S = F_3 + F_6 + F_9 + ... + F_n

2 S = 2 F 3 + 2 F 6 + 2 F 9 + . . . + 2 F n 2S = 2F_3 + 2F_6 + 2F_9 + ... + 2F_n

2 F n = F n − 2 + F n − 1 + F n 2F_n = F_{n-2} + F_{n-1} + F_n

2 S = F 1 + F 2 + F 3 + F 4 + . . . + F n 2S = F_1 + F_2 + F_3 + F_4 + ... + F_n

S = F n + 2 − 1 2 S = \frac{F_{n+2} - 1}{2}

Therefore the denominator is equal to:

F 4303 − 1 2 + F 4304 − 1 2 = F 4305 − 2 2 \dfrac{F_{4303}-1}{2} + \dfrac{F_{4304}-1}{2} = \dfrac{F_{4305}-2}{2}

Therefore,

2 F 4305 − 4 F 4305 − 2 2 = 4 \dfrac{2F_{4305}-4}{\frac{F_{4305}-2}{2}} = 4

0 Helpful 0 Interesting 1 Brilliant 1 Confused

@Braden Dean, your problem statement is wrong. By convention F 0 = 0 F_0 = 0 , F 1 = 1 F_1 = 1 , and F 2 = 1 F_2 = 1 , ... All your summation should start with ∑ i = 1 \sum_\red{i=1} instead i = 0 i=0 .

Chew-Seong Cheong - 6 months, 2 weeks ago

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Thank you! Fixed.

Braden Dean - 6 months, 2 weeks ago

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Still incorrect. The upper limits 4302 4302 , 2151 2151 , and 1434 1434 were correct.

Chew-Seong Cheong - 6 months, 2 weeks ago

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