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The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8, 13, ... .(Start from the 3rd term, each term is the sum of the two previous terms). Let $F_n$ be the $n$th term of this sequence. $S$ is defined as $S=\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$ Calculate the value of $S$

I have no idea how to solve this, hints aswell as solutions would be appreciated

Taken from the 2013 AITMO

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5 Answers 5

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Replace $F_n$ with $F_{n-1}+F_{n-2}$, now you have two series that are both similar to $S$

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    $\begingroup$ To expand on this, show that $S(x)=\sum_{k=0}^\infty F_kx^k$ is given by $S(x)-xS(x)-x^2S(x)=1$ for all small complex numbers $x$. $\endgroup$ Oct 3, 2019 at 8:53
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Hint: if you know that the generating function for the Fibonacci sequence is:

$\displaystyle \sum_{n=0}^\infty F_nx^n = \frac{x}{1-x-x^2}$

then you can substitute $x=\frac 1 2$ and you immediately have

$\displaystyle \sum_{n=0}^\infty \frac{F_n}{2^n} = \frac{\frac 1 2}{1-\frac 1 2 -\frac 1 4} =2 $

So to answer questions like this quickly, you should learn about generating functions.

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  • $\begingroup$ I think the OP looks for the steps towards a solution ; the knowledge of the generating function is no pre-knowledge for the OP, I think. $\endgroup$ Oct 3, 2019 at 11:24
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    $\begingroup$ @GottfriedHelms So this might prompt the OP to learn about generating functions ... I have edited my answer to make this clear. $\endgroup$
    – gandalf61
    Oct 3, 2019 at 12:17
  • $\begingroup$ Very well! Upvoted. $\endgroup$ Oct 3, 2019 at 13:27
  • $\begingroup$ +1 for using a generating function. Readers not familiar with generating functions might be interested in this question and its answers: math.stackexchange.com/questions/3142386/… $\endgroup$
    – awkward
    Oct 3, 2019 at 17:48
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Hint: $$F_n{={\frac {\Phi ^{n}-\Psi ^{n}}{\sqrt {5}}}={\frac {1}{\sqrt {5}}}\left(\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right)}$$

$$S = \sum_{n=1}^{\infty}\dfrac{F_n}{2^n}=\dfrac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left[\left(\dfrac{\Phi}{2}\right)^n- \left(\dfrac{\Psi}{2}\right)^n\right]$$

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  • $\begingroup$ The sign in the middle should be $-$. $\endgroup$
    – Z Ahmed
    Oct 3, 2019 at 9:08
  • $\begingroup$ @Dr Zafar Ahmend DSc Thank you for spotting the typo. $\endgroup$ Oct 7, 2019 at 18:45
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$$F_n=\frac{a^n-b^n}{\sqrt{5}},~ a+b=1, ~ab=-1,~ a,b=\frac{1\pm\sqrt{5}}{2}.$$ The required sim $$ s=\sum_{n=1}^{\infty} \frac{F_n}{2^n}= \frac{1}{\sqrt{5}} \sum_{n=1}^{\infty} \left ( \frac{a^n}{2^n}-\frac{b^n}{2^n} \right) =\frac{1}{\sqrt{5}}\left(\frac{a}{2-a}-\frac{b}{2-b}\right)= \frac{1}{\sqrt{5}}~\frac{2(a-b)}{4-2(a+b)+ab}=\frac{2 \sqrt{5}}{\sqrt{5}}=2.$$

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To expand on the hint of @empy2:

$$\begin{array}{rlll} S(z) &= 1 + & 1z +& 2z^2 + 3z^3 + 5z^4 + ... \\\ z S(z) &= & 1z +& 1z^2 + 2z^3 + 3z^4 + 5z^5 + ... \\\ S(z)-zS(z)-1 &= && 1z^2 + 1z^3 + 2z^4 + 3z^5 \\\ \end{array} \\\ \begin{array}{rlll} \hline S(z)-zS(z)-1 &= z^2 S(z) &\qquad & \phantom{sdfsdfsdfsdfs} \\\ S(z)(1-z-z^2)& =1 \\\ S(z) &= 1/(1-z-z^2) \end{array} $$ Now insert $1/2$ for $z$ and compute $1/2 S(1/2)$

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