Fraction

Calculus Level 3

1 + 1 2 + 1 1 + 1 2 + 1 1 + ⋱ \large 1 + \cfrac{1}{2 + \cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1 + \ddots}}}}

If the infinitely nested fraction above is equal to a + b c \dfrac{a+\sqrt b}c , where a , b a,b and c c are positive integers with b b square-free, find a + b + c a+b+c .


The answer is 6.

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Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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1 solution

Jason Chrysoprase
May 21, 2016

x = 1 + 1 2 + 1 1 + 1 2 + 1 1 + ⋱ = 1 + 1 2 + 1 x = 1 + 1 2 x + 1 x = 1 + x 2 x + 1 = 2 x + 1 + x 2 x + 1 3 x + 1 2 x + 1 = x 3 x + 1 = 2 x 2 + x − 2 x 2 + 2 x + 1 = 0 x = − ( 2 ) − 2 2 − 4 ( − 2 × 1 ) 2 × − 1 ( We use minus to get positive answer ) x = − 2 − 12 − 4 x = 1 + 3 2 \large x = 1 + \frac{1}{2 + \frac{1}{1+\frac{1}{2+\frac{1}{1 + \ddots}}}} \\ \large = 1 + \frac{1}{2 + \frac{1}{x}} \\ \large = 1 + \frac{1}{\frac{2x +1}{x} } \\ \large = 1 + \frac{x}{2x+1} \\ \large = \frac{2x + 1 + x}{2x+1} \\ \large \frac{3x +1}{2x +1} = x \\ \large 3x+1 = 2x^2 + x \\ \large -2x^2 + 2x + 1 = 0 \\ \large x = \frac{-(2)-\sqrt{2^2 -4(-2 \times 1)}}{2\times -1} \color{#D61F06}{\text{( We use minus to get positive answer )}} \\ x= \frac{-2-\sqrt{12}}{-4} \\ \large x = \frac{1+\sqrt{3}}{2}

a + b + c = 1 + 3 + 2 = 6 \Huge a +b+c = 1 + 3 +2 = \color{#D61F06}{\boxed{6}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

I always ask: How do we know that it converges?

Otto Bretscher - 5 years ago

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Hmm, i'm not master of calculus. I know diverges and converges but don't know how it works

Jason Chrysoprase - 5 years ago

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Maybe somebody will address the issue. If not, I guess I will have to do it since I brought it up ;)

Otto Bretscher - 5 years ago

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@Otto Bretscher – Tag someone that is good in calculus, that might help

Jason Chrysoprase - 5 years ago

This problem is 2013 Olympiad Algebra Problem

Jason Chrysoprase - 5 years ago

Nice problem & a nice solution ! +1 !

Rishabh Tiwari - 5 years ago

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thx ;), yours too

Jason Chrysoprase - 5 years ago

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