The Fibonacci sequence is given by \( F_1 = 1\), \(F_2 = 1\), and \( F_{n+1} = F_n + F_{n-1} \). What is \[ \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } ?\]

My work :

\[ \begin{align} \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } & = \sum_{n=4}^\infty \frac{ 1 } { F_{n-3} F_{n+3} } \times \frac{F_{n+3} - F_{n-3}}{4F_{n}}\\ & = \frac{1}{4} {\left( \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n-3} } - \frac{ 1 } { F_{n} F_{n+3} }\right)}\\ & = \frac{1}{4} {\left( \frac{1}{3} + \frac{1}{5} + \frac{1}{16} + 2 \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n+3} }\right)}\\ & = \frac{143}{960} + \frac{1}{2} \sum_{n=4}^\infty \frac{ 1 } { F_{n} F_{n+3} }\\ \end{align} \]

Any help will make me happy :)

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## Comments

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TopNewestJason,

subtractthose two sums\(\displaystyle \sum _{ n=4 }^{ \infty }{ \dfrac { 1 }{ { F }_{ n }{ F }_{ n+3 } } } - \displaystyle \sum _{ n=4 }^{ \infty }{ \dfrac { 1 }{ { F }_{ n }{ F }_{ n+3 } } } =0\)

and then you already have your answer!

\(= \dfrac{143}{960}+0\)

Interesting excursion into Fibonnaci identities.

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Which two sum ? Which line ?

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Hold on

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

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

From this, you should get

\( = \dfrac{1}{4} \left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{16}+ \displaystyle \sum_{n=4}^\infty \dfrac{ 1 } { F_{n+3} F_{n} } - \dfrac{ 1 } { F_{n} F_{n+3} }\right) \)

Get it now?

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