Help: Algebra (Fibonacci Numbers)

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 :

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

Any help will make me happy :)

SOLVED

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Comments

Jason, subtract those 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.

Michael Mendrin - 4 years, 4 months ago

Which two sum ? Which line ?

Jason Chrysoprase - 4 years, 4 months ago

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?

Michael Mendrin - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$