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# 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
8 months, 3 weeks ago

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Jason, subtract those two sums

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

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

Interesting excursion into Fibonnaci identities.

- 8 months, 3 weeks ago

Which two sum ? Which line ?

- 8 months, 3 weeks 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?

- 8 months, 3 weeks ago

Comment deleted 8 months ago

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.

- 8 months, 3 weeks ago

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 $\frac{143}{960} + \frac{1}{4} \left( \sum_{n=4}^{\infty} \frac{1}{F_n F_{n+3}} - \frac{1}{F_n F_{n+3}} \right)$

- 8 months, 3 weeks ago

That is right!

Boy you were so close.

- 8 months, 3 weeks ago

Owh thank you, i'm so blind :v

- 8 months, 3 weeks ago

I am trying to explain the details to you right now

- 8 months, 3 weeks ago