Help: Algebra (Fibonacci Numbers)

The Fibonacci sequence is given by F1=1 F_1 = 1, F2=1F_2 = 1, and Fn+1=Fn+Fn1 F_{n+1} = F_n + F_{n-1} . What is n=41Fn3Fn+3? \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } ?

My work :

n=41Fn3Fn+3=n=41Fn3Fn+3×Fn+3Fn34Fn=14(n=41FnFn31FnFn+3)=14(13+15+116+2n=41FnFn+3)=143960+12n=41FnFn+3 \begin{aligned} \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } & = \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } \times \frac{F_{n+3} - F_{n-3}}{4F_{n}}\\ & = \frac{1}{4} {\left( \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n-3} } - \frac{ 1 } { F_{n} F_{n+3} }\right)}\\ & = \frac{1}{4} {\left( \frac{1}{3} + \frac{1}{5} + \frac{1}{16} + 2 \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n+3} }\right)}\\ & = \frac{143}{960} + \frac{1}{2} \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n+3} }\\ \end{aligned}

Any help will make me happy :)

SOLVED

Note by Jason Chrysoprase
2 years, 9 months ago

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1 vote

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Jason, subtract those two sums

n=41FnFn+3n=41FnFn+3=0\displaystyle \sum _{ n=4 }^{ \infty }{ \dfrac { 1 }{ { F }_{ n }{ F }_{ n+3 } } } - \displaystyle \sum _{ n=4 }^{ \infty }{ \dfrac { 1 }{ { F }_{ n }{ F }_{ n+3 } } } =0

and then you already have your answer!

=143960+0= \dfrac{143}{960}+0

Interesting excursion into Fibonnaci identities.


Michael Mendrin - 2 years, 9 months ago

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Which two sum ? Which line ?

Jason Chrysoprase - 2 years, 9 months ago

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Hold on

Okay, remember this line? You're doing a subtraction. How did you end up with an addition?

=14(n=41FnFn31FnFn+3) = \dfrac{1}{4} \left( \displaystyle \sum_{n=4}^\infty \dfrac{ 1 } { F_{n} F_{n-3} } - \dfrac{ 1 } { F_{n} F_{n+3} }\right)

From this, you should get

=14(13+15+116+n=41Fn+3Fn1FnFn+3) = \dfrac{1}{4} \left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{16}+ \displaystyle \sum_{n=4}^\infty \dfrac{ 1 } { F_{n+3} F_{n} } - \dfrac{ 1 } { F_{n} F_{n+3} }\right)

Get it now?

Michael Mendrin - 2 years, 9 months ago

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