Waste less time on Facebook — follow Brilliant.
×

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 :

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

Any help will make me happy :)

SOLVED

Note by Jason Chrysoprase
6 months, 3 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 · 6 months, 3 weeks ago

Log in to reply

@Michael Mendrin Which two sum ? Which line ? Jason Chrysoprase · 6 months, 3 weeks ago

Log in to reply

@Jason Chrysoprase 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 · 6 months, 3 weeks ago

Log in to reply

Comment deleted 6 months ago

Log in to reply

@Jason Chrysoprase You are already taking the difference between two [different] series. How did you come to decide to add two series? I tell you, \(\dfrac{143}{960}\) is the numerical answer you are looking for. Michael Mendrin · 6 months, 3 weeks ago

Log in to reply

@Michael Mendrin I only need the prove, answer without prove is nothing for me

So what you are trying to say is i have done something wrong in my work ? In line 3 from above

So that i should end up with \[\frac{143}{960} + \frac{1}{4} \left( \sum_{n=4}^{\infty} \frac{1}{F_n F_{n+3}} - \frac{1}{F_n F_{n+3}} \right)\] Jason Chrysoprase · 6 months, 3 weeks ago

Log in to reply

@Jason Chrysoprase That is right!

Boy you were so close. Michael Mendrin · 6 months, 3 weeks ago

Log in to reply

@Michael Mendrin Owh thank you, i'm so blind :v Jason Chrysoprase · 6 months, 3 weeks ago

Log in to reply

@Jason Chrysoprase I am trying to explain the details to you right now Michael Mendrin · 6 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...