In my book they calculated the $28$th Fibonacci number and said $F_{28} = 3 \times 13 \times 19 \times 281 = 317811$. This made me wonder if there was an easier way to find the $28$th Fibonacci number than by doing it by hand.
52907
I was reading up on the Fibonacci Sequence,  $\text {{1,1,2,3,5,8,13,....}}$ when I've noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position.  Is there a way to do this? If so, how are we able to apply these formulas to arrays?
92526
How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^) I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were: $$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$  $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
27798
Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$   My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?
120670
While solving , I discovered that there is a relationship between the Fibonacci sequence and the golden ratio.  After I got the correct answer via brute force, I discovered this relationship.  One of the posters said this:     The nth Fibonacci number is $[\phi^n / \sqrt{5}\;]$,  where the brackets   denote "nearest integer".       So we need $\phi^n/\sqrt{5} &gt; 10^{999}$   $n \log(\phi) - \log{5}/2 &gt; 999 \log(10)$   $n \log(\phi) &gt; 999 \log(10) + \log(5)/2$   $n &gt; (999 \log(10) + \log(5) / 2) / \log(\phi)$       A handheld calculator shows the right hand side to be 4781.8593, so   4782 is the first integer n with the desired property.   I can't quite make sense of this.  How does the poster know that the $n$-th term is $\phi^n / \sqrt{5}$?  Can you also walk me through the operations?
72678
Where $n \in \mathbb{N}$   and  $$ F_n = \begin{cases} 0                 &amp; \text{ if } n = 0 \\ 1                 &amp; \text{ if } n = 1 \\ F_{n-1} + F_{n-2} &amp; \text{ if } n &gt; 1 \end{cases} $$  This is basically describing the famous Fibonacci sequence.  If we try the base case with 1, it works (for $0$ I am not sure...)  When $n = 1$  $2 \cdot0 - 1^2 = (-1)^1 = -1$  For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.  Now, for the inductive step, we try to prove for $n+1$, so for  $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.  Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even).   Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).  Or simply: $$ (-1)^n = -(-1)^{n+1} $$ $$ 1 \cdot (-1)^n = -(-1)\cdot(-1)^n $$ $$ 1 = -(-1) $$ Which is true.  I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...
106823
Let $\{ F_n\}$ be the sequence of numbers defined by $F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}$ for $n \geq 2$. Let $f_n$ be the remainder left when $F_n$ is divided by $5$. Then $f_{2000}$ equals (A) $0$ $~~~~~~~~~~~~~~~~$ (B) $1$   $~~~~~~~~~~~~~~~~$ (C) $2$$~~~~~~~~~~~~~~~~$ (D) $3$  I found that $F(1)=1$, $F(2)=1$, $F(3)=2$, $F(4)=3$, $F(5)=5$, $F(6)=8$, $F(7)=13$, $F(8)=21$, $F(9)=34$, and $F(10)=55$. But I need a systematic pattern to find $F_n$.
22974
May someone help me? I am trying to use induction to prove that the formula for finding the $n$-th term of the Fibonacci sequence is:   $$F_n=\frac{1}{\sqrt{5}}⋅\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}⋅\left(\frac{1-\sqrt{5}}{2}\right)^n.$$  I tried to put $n=1$ into the equation and prove that if $n=1$ works then $n=2$ works and it should work for any number, but it didn't work. I need to prove that this formula gives the $n$th Fibonacci number.
120024
Make and prove a conjecture about when the Fibonacci sequence, $F_n$, is divisible by $7$.   I've realized it's when $n$ is a multiple of $8$. I just don't know how to go about proving it.
39515
Let $f_n$ be the Fibonacci numbers, defined by $f_0 = 0$ and $f_1 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for all $n \ge 2$.  Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer.  Just some homework help. Need to prove. Thank you in advance.
122378
I stumbled upon the following in a book:     till 400 all even numbers will be divisible by 2 ( 200 even numbers)   remaining 200 odd numbers  1 3 5 7 9 ..... 399 200/3 = 67 will be   divisible by 3, as we have discarded all even numbers and out of   remaining numbers 2nd number is 3 then 1 + 198/3 = 67 numbers are   divisible by 3      now we have discarded all multiples of 2 and 3 remaining numbers will   be  200 - 67 = 133;   1 5 7 11 13 ... so on out of these 133 numbers 2nd   number is 5 so numbers divisible by 5 will be 1 + 131/5 = 1+26 = 27      now remaining numbers = 133 - 27 = 106; 1 7 11 13 .. so on again 2nd   number is 7 , so numbers divisible by 7 = 1 + 104/7 = 1+14 = 15      total numbers from 1 - 400 divisible by any of 2 3 5 7 = 200+67+27+15   = 309   But I can't seem to understand how the following is derived? :     as we have discarded all even numbers and out of remaining numbers 2nd   number is 3 then 1 + 198/3 = 67 numbers are divisible by 3   How does the result come out to be "1 + 198/3" ?
66720
I have fibonacci numbers defined as such:  $$  f(n) = f(n-1) + f(n-2)$$with$$    f(0) = 0 $$$$f(1) = 1$$  I have to prove that $$ F(n) \geq 1.5^{n-1}, n \geq 6$$  Base Case:  $$ f(6) = 8 \geq (1.5)^5 = 7.6 $$  Base holds  Inductive hypothesis $$  k \geq 6 $$$$ f(k) \geq (1.5)^{k-1} $$  Inductive Step  Here is how I have started, any hints or pointers on how to continue are appreciated  $$ f(k-1) \geq (1.5)^{k+1-1}$$ $$ f(k) + f(k-1) = (1.5)^{k-1} + (1.5)^{k-2}$$ now I think where I want to go from here is to expand the right side of this equation out, but I am unsure how to do this.
92669
For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?  This problem is taken from the today test:   I think we can consider $$f(x)=\dfrac{\ln{x}}{x}$$  or perhaps other methods, so I am looking forward to seeing other methods to solve this problem. Thank you.
53762
The definition of a Fibonacci number is as follows:  $$F_0=0\\F_1=1\\F_n=F_{n-1}+F_{n-2}\text{ for } n\geq 2$$   Prove the given property of the Fibonacci numbers for all n greater than or equal to 1. $$F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$$  I am pretty sure I should use weak induction to solve this. My professor got me used to solving it in the following format, which I would like to use because it help me map everything out...  This is what I have so far:  Base Case: Solve for $F_0$ and $F_1$ for the following function: $F_nF_{n+1}$.  Inductive Hypothesis:  What I need to show: I need to show $F_{n+1}F_{n+1+1}$ will satisfy the given property. Proof Proper:  (didn't get to it yet)  Any intro. tips and pointers?
85502
[question:]      Prove by induction that the i th Fibonacci number satisfies the equality $$F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}$$where $\phi$ is the golden ratio and $\hat{\phi}$ is its conjugate. [end]  I've had multiple attempts at this, the most fruitful being what follows, though it is incorrect, and I cannot figure out where I am going wrong:  [my answer:] What follows is an incorrect approach, after which comes the approach that I should have done  Inductive Hypothesis: $$F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}, i\in\mathbb{N}  $$   (I include 0) Base case: $$F_0=\frac{\phi^0-\hat{\phi^0}}{\sqrt5}=0$$  Proof: $$\begin{eqnarray*} F_{i+1}&amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ &amp;=&amp;\frac{\phi*\phi^i-\hat{\phi}*\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{1+\sqrt5}{2}*\phi^i-\frac{1-\sqrt5}{2}*\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{\phi^i+\sqrt5\phi^i}{2}-\frac{\hat{\phi^i}-\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &amp;=&amp;\frac{\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &amp;=&amp;\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2*\sqrt5}\\ &amp;=&amp;\frac{1}{2}\left(\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{\sqrt5}\right)\\ &amp;=&amp;\frac{1}{2}\left(\frac{\sqrt5\phi^i}{\sqrt5} + \frac{\sqrt5\hat{\phi^i}}{\sqrt5} +\frac{\phi^i-\hat{\phi^i}}{\sqrt5}\right)\\ &amp;=&amp;\frac{1}{2}\left(\phi^i+\hat{\phi^i}+F_{i}\right)\text{   by inductive hypothesis}\\ &amp;=&amp;\frac{1}{2}\left(\sqrt5*...\right)\\ \end{eqnarray*}$$ Actually I just saw my error in that line (I eventually multiplied by $\frac{\sqrt5}{\sqrt5}$ and  substituted for $F_i$, but I see that the conjugates are added there, not subtracted).  Correct approach: $$\begin{eqnarray*} F_{i+1}&amp;=&amp;F_{i} + F_{i-1}\\ &amp;=&amp;\frac{\phi^i-\hat{\phi^i}}{\sqrt5}+\frac{\phi^{i-1}-\hat{\phi^{i-1}}}{\sqrt5}\\ &amp;=&amp;\frac{\left(\phi+\hat{\phi}\right)\left(\phi^i-\hat{\phi^i}\right)-\phi\hat{\phi}\left(\phi^{i-1}-\hat{\phi^{i-1}}\right)}{\sqrt5}\text{ (see answer for why this works)}\\ &amp;=&amp;\frac{\phi^{i+1}-\phi\hat{\phi^i}+\hat{\phi}\phi^i-\hat{\phi^{i+1}}-\phi^i\hat{\phi}+\phi\hat{\phi^i}}{\sqrt5}\\ &amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}-\phi\hat{\phi^i}+\phi\hat{\phi^i}-\hat{\phi}\phi^i+\hat{\phi}\phi^i}{\sqrt5}\\ &amp;=&amp;\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ \text{Q.E.D., punk problem.} \end{eqnarray*}$$  From what I searched on the web, there is no available source proving this of the Fibonacci sequence - all inductive hypothesis proofs prove the actual sequence with induction.   Bill, it's always easy to forget about the inherent properties of special numbers... I didn't even realize $\phi$ + $\hat{\phi}$ = 1 in your first hint, as well as the multiplication, until you pointed it out. Alex, I'm sure that going through the problem that way will produce the result, also, but with Bill's, it's much faster.    Thanks for the help.
33620
I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.  Here's my method:  Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$  $$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$  I just felt like this was sort of a messy way to find out that the last two digits are 43.
116892
In The Fibonacci sequence which is defined as $$ F_n=F_{n-1}+F_{n-2}, $$ lets say we have the number $p$ which is an odd prime.   Prove that:   $F_{p-1} + F_{p+1} -1$ Is divisible by $p$. Prove that for any given $n$ real positive integer:  $F_{p^{n+1}-1} + F_{p^{n+1}+1} -(F_{p^{n}-1} + F_{p^{n}+1})$ Is divisible by $p^{n+1}$   How to prove this?
9434
Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step.  Identity: $$f_n^2 + f_{n+1}^2 = f_{2n+1}$$  I get to:  $$f_{n+1}^2 + f_{n+2}^2$$  Should I replace $f_{n+2}$ using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step?  Thanks!