This is the continuation of the note by Fahim Shariar Shakkhor

The \(\text{Fibonacci Sequence}\) is a sequence of integers where the first \(\displaystyle 2\) term are \(\displaystyle 0\) and \(\displaystyle 1\), and each subsequent term is the sum of the previous two numbers.

\[ 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377, ... \]

The \(n\)-th Fibonacci number is denoted by \(F_n\) and the sequence is defined by the recurrence relation

\[\large F_n = F_{n-1} + F_{n-2}.\]

Here, \(F_0 = 0,\enspace F_1 = 1,\enspace F_2 = F_1 + F_0 = 1\) etc. \[\]

\[ \large F_n = \dfrac{ \left( \dfrac{1+\sqrt{5}}{2} \right)^{n}- \left(\dfrac{1-\sqrt{5}}{2} \right)^{n} }{\sqrt{5}} \]

\(\text{Problem 1:}~\)(By - Jake Lai) Prove that for all - \(\large |x| < \phi^{-1}\)

\(\large \displaystyle \dfrac{x}{1-x-x^{2}} = \sum_{k=0}^{\infty} F_{k}x^{k}\)

\(\text{Problem 2:} ~~ \large \displaystyle \sum_{i=0}^{n} \dfrac{F_{i}}{i}\)

\(\text{Problem 3:} ~~ \large \displaystyle \sum_{i=1}^{n} \frac{1}{F_{i+1}\times F_{i}}\)

\(\text{Problem 4:} ~~\large \displaystyle \sum_{i=1}^{n} \frac{1}{F_{i}}\)

\(\text{Problem 5:}\) By Azhaghu Roopesh M

\(\large \displaystyle \sum_{p=0}^{n-1} \binom{n-p}{p} = F_{n+1}\)

## Comments

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– Megh Choksi · 2 years, 1 month ago

I am writing it as a question , so you can delete it from here.Log in to reply

EDIT:- This is for Q5 – Siddhartha Srivastava · 2 years, 1 month ago

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Problem 4:

Given below is a Java Code (Netbeans IDE)

Sorry, I couldn't get the code arranged properly using LaTex as @Brock Brown does, so I had to upload a pic .

@megh choksi , I don't think an accurate partial sum can be developed, so I just let the upper bound be a very large number since I guessed that it will converge(yes it does !) .

Can you write a Mathematical solution ? Take the \( Lim_{n \rightarrow \infty}\) – Azhaghu Roopesh M · 2 years, 1 month ago

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@megh choksi Also remember \(F_0 = 0\) so change the lower bounds of the summations wherever necessary :) – Azhaghu Roopesh M · 2 years, 1 month ago

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– Megh Choksi · 2 years, 1 month ago

Thanks editedLog in to reply