Infinite Fibonacci Fractions

Algebra Level 3
  1. Let a/b (where both a and b are positive integers ) equal 1/5+1/25+2/125+...+F(n)/5^n, where F(n) is the nth Fibonacci number, defined by F(1) = F(2) = 1 and F(n)= F(n-1) + F(n-2) for all integers n>2. Find 10a+b.


The answer is 69.

Daniel Liu by Daniel Liu , 21, USA

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1 solution

Nisshith Sharma
Mar 5, 2014

a b = 1 5 + 1 5 2 + 2 5 3 + 3 5 4 + 5 5 5 + 8 5 6 + 13 5 7 . . . . ∞ a 5 b = 1 5 2 + 1 5 3 + 2 5 4 + 3 5 5 + 5 5 6 + 8 5 7 . . . . ∞ a b − a 5 b = 1 5 + 0 + 1 5 3 + 1 5 4 + 2 5 5 + 3 5 6 + 5 5 7 . . . . ∞ T h e r e f o r e , a b − a 5 b = 1 5 + a 25 b S o l v i n g t h i s w e g e t , a b = 5 19 H e n c e , 10 a + b = 69. \quad \quad \quad \quad \quad \quad \frac { a }{ b } =\quad \frac { 1 }{ 5 } +\frac { 1 }{ { 5 }^{ 2 } } +\frac { 2 }{ { 5 }^{ 3 } } +\frac { 3 }{ { 5 }^{ 4 } } +\frac { 5 }{ { 5 }^{ 5 } } +\frac { 8 }{ { 5 }^{ 6 } } +\frac { 13 }{ { 5 }^{ 7 } } ....\infty \\ \quad \quad \quad \quad \quad \frac { a }{ 5b } =\quad \quad \quad \quad \quad \frac { 1 }{ { 5 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 3 } } +\frac { 2 }{ { 5 }^{ 4 } } +\frac { 3 }{ { 5 }^{ 5 } } +\frac { 5 }{ { 5 }^{ 6 } } +\frac { 8 }{ { 5 }^{ 7 } } ....\infty \\ \\ \frac { a }{ b } -\frac { a }{ 5b } \quad =\quad \frac { 1 }{ 5 } +\quad 0+\frac { 1 }{ { 5 }^{ 3 } } +\frac { 1 }{ { 5 }^{ 4 } } +\frac { 2 }{ { 5 }^{ 5 } } +\frac { 3 }{ { 5 }^{ 6 } } +\frac { 5 }{ { 5 }^{ 7 } } ....\infty \\ Therefore,\\ \frac { a }{ b } -\frac { a }{ 5b } \quad =\quad \frac { 1 }{ 5 } +\quad \frac { a }{ 25b } \\ \\ Solving\quad this\quad we\quad get,\\ \frac { a }{ b } \quad =\quad \frac { 5 }{ 19 } \quad \\ Hence,\quad 10a\quad +\quad b\quad =\quad 69.

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wahh,,,,,,,can you teach me ?? please

Quennie Joy - 7 years, 3 months ago

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