Fibonacci Decimals

Algebra Level 4

1 1 0 0 + 1 1 0 1 + 2 1 0 2 + 3 1 0 3 + 5 1 0 4 + 8 1 0 5 + 13 1 0 6 + ⋯ \displaystyle \frac{1}{10^0} + \frac{1}{10^1} + \frac{2}{10^2} + \frac{3}{10^3} + \frac{5}{10^4} + \frac{8}{10^5} + \frac{13}{10^6} + \cdots

If the value of the sum above can be expressed as m n \dfrac mn for positive coprime integers m m and n n , find n + 1 m \dfrac{n + 1}{\sqrt{m}} .


The answer is 9.

Zach Abueg by Zach Abueg , 22, USA

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3 solutions

Relevant wiki: Fibonacci Sequence

S = 1 1 0 0 + 1 1 0 1 + 2 1 0 2 + 3 1 0 3 + 5 1 0 4 + 8 1 0 5 + 13 1 0 6 + ⋯ = ∑ n = 1 ∞ F n 1 0 n − 1 F n is the n th Fibonacci number. = 10 ∑ n = 0 ∞ F n 1 0 n = 10 ( F 0 + F 1 10 + ∑ n = 2 ∞ F n 1 0 n ) = 10 ( 1 10 + ∑ n = 2 ∞ F n − 1 + F n − 2 1 0 n ) = 10 ( 1 10 + ∑ n = 2 ∞ F n − 1 1 0 n + ∑ n = 2 ∞ F n − 2 1 0 n ) = 10 ( 1 10 + 1 10 ∑ n = 0 ∞ F n 1 0 n + 1 100 ∑ n = 0 ∞ F n 1 0 n ) = 1 + 1 10 S + 1 100 S \begin{aligned} S & = \small \frac 1{10^0} + \frac 1{10^1} + \frac 2{10^2} + \frac 3{10^3} + \frac 5{10^4} + \frac 8{10^5} + \frac {13}{10^6} + \cdots \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac {\color{#3D99F6}F_n}{10^{n-1}} & \small \color{#3D99F6} F_n \text{ is the }n \text{th Fibonacci number.} \\ & = 10 \sum_{\color{#D61F06}n=0}^\infty \frac {F_n}{10^n} \\ & = 10 \left(F_0 + \frac {F_1}{10} + \sum_{\color{#3D99F6}n=2}^\infty \frac {F_n}{10^n} \right) \\ & = 10 \left(\frac 1{10} + \sum_{\color{#3D99F6}n=2}^\infty \frac {F_{n-1}+F_{n-2}}{10^n} \right) \\ & = 10 \left(\frac 1{10} + \sum_{\color{#3D99F6}n=2}^\infty \frac {F_{n-1}}{10^n} + \sum_{\color{#3D99F6}n=2}^\infty \frac {F_{n-2}}{10^n}\right) \\ & = 10 \left(\frac 1{10} + \frac 1{10} \sum_{\color{#D61F06}n=0}^\infty \frac {F_n}{10^n} + \frac 1{100} \sum_{\color{#D61F06}n=0} ^\infty \frac {F_n}{10^n}\right) \\ & = 1 + \frac 1{10} S + \frac 1{100} S \end{aligned}

⟹ 89 100 S = 1 S = 100 89 ⟹ n + 1 m = 89 + 1 100 = 9 \begin{aligned} \implies \frac {89}{100} S & = 1 \\ S & = \frac {100}{89} \\ \implies \frac {n+1}{\sqrt m} & = \frac {89+1}{\sqrt{100}} = \boxed{9} \end{aligned}

7 Helpful 1 Interesting 2 Brilliant 0 Confused

I somehow deleted my solution and now I can't post it again, so I'll just write it here:

Let the sum be S:

S = 1 + 1 10 + 2 1 0 2 + 3 1 0 3 + 5 1 0 4 + 8 1 0 5 + . . . S=1+\frac{1}{10}+\frac{2}{10^{2}}+\frac{3}{10^{3}}+\frac{5}{10^{4}}+\frac{8}{10^{5}}+...

1 10 S = 1 10 + 1 1 0 2 + 2 1 0 3 + 3 1 0 4 + 5 1 0 5 + . . . \frac{1}{10}S=\frac{1}{10}+\frac{1}{10^{2}}+\frac{2}{10^{3}}+\frac{3}{10^{4}}+\frac{5}{10^{5}}+...

⇒ 9 10 S = 1 + 1 10 − 1 10 ⏟ + 2 1 0 2 − 1 1 0 2 ⏟ + 3 1 0 3 − 2 1 0 3 ⏟ + 5 1 0 4 − 3 1 0 4 ⏟ + 8 1 0 5 − 5 1 0 5 ⏟ + . . . = 1 + 1 1 0 2 + 1 1 0 3 + 2 1 0 4 + 3 1 0 5 + 5 1 0 6 + . . . \begin{aligned} \Rightarrow \frac{9}{10}S & =1+\underbrace{\frac{1}{10}-\frac{1}{10}}+\underbrace{\frac{2}{10^{2}}-\frac{1}{10^{2}}}+\underbrace{\frac{3}{10^{3}}-\frac{2}{10^{3}}}+\underbrace{\frac{5}{10^{4}}-\frac{3}{10^{4}}}+\underbrace{\frac{8}{10^{5}}-\frac{5}{10^{5}}}+...\\ & =1+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\frac{2}{10^{4}}+\frac{3}{10^{5}}+\frac{5}{10^{6}}+...\end{aligned}

At this moment note that 1 1 0 2 + 1 1 0 3 + 2 1 0 4 + 3 1 0 5 + 5 1 0 6 + . . . = S 1 0 2 \frac{1}{10^{2}}+\frac{1}{10^{3}}+\frac{2}{10^{4}}+\frac{3}{10^{5}}+\frac{5}{10^{6}}+...=\frac{S}{10^{2}} .

⇒ 9 10 S = 1 + S 1 0 2 ⇒ 89 100 S = 1 ⇒ S = 100 89 ⇒ m = 100 , n = 89 \begin{aligned} & \Rightarrow \frac{9}{10}S=1+\frac{S}{10^{2}} \\ & \Rightarrow \frac{89}{100}S=1 \\ & \Rightarrow S=\frac{100}{89} \\ & \Rightarrow m=100, n=89 \end{aligned}

Now, n + 1 m = 90 10 = 9 \frac{n+1}{\sqrt{m}}=\frac{90}{10}=\boxed{9} .

Uros Stojkovic - 3 years, 10 months ago
Marco Brezzi
Aug 8, 2017

Using Binet's Formula

F n = ( 1 + 5 2 ) n − ( 1 − 5 2 ) n 5 F_n=\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}

The required sum is

S = ∑ n = 0 ∞ F n 1 0 n − 1 = 1 5 ∑ n = 0 ∞ 1 1 0 n − 1 ⋅ [ ( 1 + 5 2 ) n − ( 1 − 5 2 ) n ] = 10 5 ∑ n = 0 ∞ ( 1 + 5 20 ) n − ( 1 − 5 20 ) n = 10 5 ( 1 1 − 1 + 5 20 + 1 1 − 1 − 5 20 ) = 100 89 \begin{aligned} S &=\sum_{n=0}^{\infty} \dfrac{F_n}{10^{n-1}}\\ & = \dfrac{1}{\sqrt{5}}\sum_{n=0}^{\infty} \dfrac{1}{10^{n-1}}\cdot \left[\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right]\\ & = \dfrac{10}{\sqrt{5}}\sum_{n=0}^{\infty} \left(\dfrac{1+\sqrt{5}}{20}\right)^n-\left(\dfrac{1-\sqrt{5}}{20}\right)^n\\ & = \dfrac{10}{\sqrt{5}}\left(\cfrac{1}{1-\frac{1+\sqrt{5}}{20}}+\cfrac{1}{1-\frac{1-\sqrt{5}}{20}}\right)=\dfrac{100}{89} \end{aligned}

Making the answer

89 + 1 100 = 9 \dfrac{89+1}{\sqrt{100}}=\boxed{9}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

let

F 0 = 1 F_0=1

F 1 = 1 F_1=1

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

P ( x ) = F 0 + F 1 x + F 2 x 2 + . . . . P(x)=F_0+F_1x+F_2x^2+....

Then we have

( 1 − x − x 2 ) P ( x ) = F 0 + ( F 1 − F 0 ) x + ( F 2 − F 1 − F 0 ) x 2 + ( F 3 − F 2 − F 1 ) x 3 + . . . (1-x-x^2)P(x)=F_0+(F_1-F_0)x+(F_2-F_1-F_0)x^2+(F_3-F_2-F_1)x^3+...

Every term except the constant is zero so we are left with

( 1 − x − x 2 ) P ( x ) = F 0 = 1 (1-x-x^2)P(x)=F_0=1

P ( x ) = 1 1 − x − x 2 P(x)=\frac{1}{1-x-x^2}

Our desired value is simply

P ( 1 / 10 ) = 100 / 89 P(1/10)=100/89

Thus

m = 100 n = 89 m=100 n=89

Giving our solution as

n + 1 100 = 90 10 = 9 \frac{n+1}{\sqrt{100}}=\frac{90}{10}=9

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice use of the Fibonacci generating function!

Zach Abueg - 3 years, 10 months ago

how did you get the 5th equation?

Vincentpaul Fadri - 3 years, 9 months ago

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