Play with Fibonacci

This is the continuation of the note by Fahim Shariar Shakkhor

The Fibonacci Sequence\text{Fibonacci Sequence} is a sequence of integers where the first 2\displaystyle 2 term are 0\displaystyle 0 and 1\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,... 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377, ...

The nn-th Fibonacci number is denoted by FnF_n and the sequence is defined by the recurrence relation

Fn=Fn1+Fn2.\large F_n = F_{n-1} + F_{n-2}.

Here, F0=0,F1=1,F2=F1+F0=1F_0 = 0,\enspace F_1 = 1,\enspace F_2 = F_1 + F_0 = 1 etc.

Fn=(1+52)n(152)n5 \large F_n = \dfrac{ \left( \dfrac{1+\sqrt{5}}{2} \right)^{n}- \left(\dfrac{1-\sqrt{5}}{2} \right)^{n} }{\sqrt{5}}


Problem 1: \text{Problem 1:}~(By - Jake Lai) Prove that for all - x<ϕ1\large |x| < \phi^{-1}

x1xx2=k=0Fkxk\large \displaystyle \dfrac{x}{1-x-x^{2}} = \sum_{k=0}^{\infty} F_{k}x^{k}

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

Problem 3:  i=1n1Fi+1×Fi\text{Problem 3:} ~~ \large \displaystyle \sum_{i=1}^{n} \frac{1}{F_{i+1}\times F_{i}}

Problem 4:  i=1n1Fi\text{Problem 4:} ~~\large \displaystyle \sum_{i=1}^{n} \frac{1}{F_{i}}

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

p=0n1(npp)=Fn+1\large \displaystyle \sum_{p=0}^{n-1} \binom{n-p}{p} = F_{n+1}

Note by U Z
4 years, 6 months 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 Limn Lim_{n \rightarrow \infty}

A Brilliant Member - 4 years, 6 months ago

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@megh choksi Also remember F0=0F_0 = 0 so change the lower bounds of the summations wherever necessary :)

A Brilliant Member - 4 years, 6 months ago

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Thanks edited

U Z - 4 years, 6 months ago

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