Middle Fibonacci Number

In general we have

$F_{n+4} = F_{n+3} + F_{n+2} = (F_{n+2} + F_{n+1}) + F_{n+2} = 2*F_{n+2} + F_{n+1}$ .

Now $F_{n+2} = F_{n+1} + F_{n} \Longrightarrow F_{n+1} = F_{n+2} - F_{n}$ .

Thus $F_{n+4} = 2*F_{n+2} + (F_{n+2} - F_{n}) = 3*F_{n+2} - F_{n}$

$\Longrightarrow F_{n+2} = \dfrac{F_{n+4} + F_{n}}{3}$ .

Plugging in $n = 24$ we have

$F_{26} = \dfrac{F_{28} + F_{24}}{3} = \dfrac{317811 + 46368}{3} = \boxed{121393}$ .

We let ${ F }_{ 24 }=a,\quad { F }_{ 25 }=b,\quad { F }_{ 26 }=a+b,\quad { F }_{ 27 }=a+2b\quad and\quad { F }_{ 28 }=2a+3b$

See what happens when we add ${ F }_{ 24 }\quad and\quad { F }_{ 28 }$ : ${ F }_{ 24 }+{ F }_{ 28 }=a+(2a+3b)=3(a+b)$

We know that $a+b={ F }_{ 26 }$

And we already know the value of ${ F }_{ 24 }\quad and\quad { F }_{ 28 }$

So substitution: $\frac { { F }_{ 24 }+{ F }_{ 28 } }{ 3 } =\frac { 46368+317811 }{ 3 } = \boxed{121393}$

By Catalan's Identity, $F_{n}^{2} = (-1)^{n+r}F_{r}^{2}+F_{n-r}F_{n+r}$ .

Letting $n = 26$ and $r = 2$ , we have $F_{26} = \sqrt{F_{2}^{2}+F_{24}F_{28}} = \sqrt{1+46368 \times 317811} = \boxed{121393}$ .

First we write those 3 identities :

${ F }_{ 28\quad }=\quad { F }_{ 27 }\quad +\quad { F }_{ 26 }\\ \\ { F }_{ 27 }\quad =\quad { F }_{ 26 }\quad +\quad { F }_{ 25 }\\ \\ { F }_{ 25 }\quad +\quad { F }_{ 24 }\quad =\quad { F }_{ 26 }\\$

After ading up, we have :

${ F }_{ 28\quad }+\quad { F }_{ 24 }\quad =\quad 3{ F }_{ 26 }\\ \\ \\ { F }_{ 26 }\quad =\quad \frac { { F }_{ 28 }+{ F }_{ 24 } }{ 3 } \\ \\ { F }_{ 26 }\quad =\quad \frac { 46368\quad +\quad 317811 }{ 3 } \quad =\boxed{121393}\quad$

In fibonacci series, as we go bigger, the ratio between 2 consecutive numbers is approximately equal to phi. So I multiplied 46368 x 1.618 x 1.618 and rounded off to get 121393

We find, by inspection, that $F_2=\frac{F_0+F_4}{3}$ and $F_3=\frac{F_1+F_5}{3}.$ Adding these equations yields $F_4=\frac{F_2+F_6}{3}$ and, by induction, $F_n=\frac{F_{n-2}+F_{n+2}}{3}$ for $n\geq2.$ For $n=26$ we find $F_{26}=\frac{F_{24}+F_{28}}{3}=121393$ .

$F_{28}=F_{27}+F_{26}\\F_{27}=F_{26}+F_{25}\\F_{26}=F_{25}+F_{24}$

Substitute $317811$ for $F_{28}$ in the first equation. Then, use the second and third equations to write the entire first equation in terms of $F_{26}$ and $F_{24}$ (note that $F_{25}=F_{26}-F_{24}$ from the third equation). Finally, substitute $46368$ for $F_{24}$ and solve for $F_{26}$ .

$317811=F_{27}+F_{26}\\317811=F_{26}+F_{25}+F_{26}\\317811=F_{26}+F_{26}-F_{24}+F_{26}\\317811=3F_{26}-46368\\3F_{26}=364179\\ \boxed{F_{26}=121393}$

If you know that Fibonacci is essentially a geometric series rounded to integers, you only have to take the geometric mean of the two numbers and observe that the result is extremely close to an integer. It must be our answer.

However, I respect the solutions here which did not require a calculator (or an extremely tedious paper calculation).

My solution is pretty much the same as the others but here it is....

We have this formula

$\color{#3D99F6}{F_{24}+F_{25}+F_{27}=F_{28}}$

Based on

$\color{#20A900}{F_{n+2}=F_{n+1}+F_n}$

Now we can manipulate the defining formula for Fibbonacci

$\color{#20A900}{F_{n+2}=F_{n+1}+F_n}$

$\color{#20A900}{\Rightarrow F_n+F_{n+2}=2F_n+F_{n+1}}$

Then we'll replace $\color{#3D99F6}{F_{25}+F_{27}}$ in the starting formula

$\color{#3D99F6}{F_{24}+}\color{#D61F06}{2F_{25}+F_{26}}\color{#3D99F6}{=F_{28}}$

Now we can form a system of equations

$\color{#3D99F6}{F_{24}+}\color{#D61F06}{2F_{25}+F_{26}}\color{#3D99F6}{=F_{28}}$ $\land$ $\color{#20A900}{F_{26}=F_{25}+F_{24}}$

Which we can solve easily

$\color{#3D99F6}{F_{24}+}\color{#D61F06}{2F_{25}+}\color{#20A900}{(F_{24}+F_{25})}\color{#3D99F6}{=F_{28}}$

$2F_{24}+3F_{25}=F_{28}$

$F_{25}=\frac{F_{28}-2F_{24}}{3}$

$F_{25}=75 025$

Now that we know $F_{25}$ , we can find out $F_{26}$

$F_{26}=F_{24}+F_{26}$

$F_{26}=75 025+46 368=121 393$

My approach was somewhat simple: We know the Fibonacci sequence changes by $\approx1.618$ . And that we can express phi as $2cos(36)$ . So, in order to find $F_{26}$ , we take $F_{24}$ and multiply by phi twice: $46386\cdot2cos(36)=75025$ And $75025\cdot2cos(36)=121393$ We can simply check our answer: $121393-75025=46386$ So the answer is $\boxed{121393}$ .

I can't believe it. This is my first level 4 Algebra solution. Am I dreaming?

My simple solution: We have $F_{24}, F_{25}, F_{26}, F_{27}, F_{28}$ .

We know all about the Fibonacci sequence, like: $F_{n} = F_{n-1} + F_{n-2}$ .

We also know the value of the first and last number, which are: $F_{24} = 46369$ and $F_{28} = 317811$ .

And for things get more clear (and easy to work with algebra), let's say that: $F_{24} = a$ , $F_{25} = b$ , $F_{26} = c$ (which we want to find it), $F_{27} = d$ and $F_{28} = e$ .

So, $a = 46369$ and $e = 312811$ .

We can assume that:

$a + b = c$ $\Longrightarrow b = c - 46368$

$b + c = d$

$c + d = e$ $\Longrightarrow d = 317811 - c$

Making some basic algebra:

$c - 46368 + c = 317811 - c$

$3c = 364179$

$c = 121393$

Which correspond to: $F_{26} = 121393$ .

Let $x_0 = F_{24}$ , $x_4 = F_{28}$ and $n = F_{26} = x_2$

Translating the formula given into our terms gives us

$x_{n+2} = x_{n+1} + x_n$

(Not really much of a difference but still needed)

This means that

$x_4 = x_3 + x_2$

This can be written in a different way by substituting

$x_4 = (n + x_1) + (n) \Rightarrow x_4 = 2n + x_1$

Since $n = x_2 = x_1 + x_0$ we can substitute that in to get

$x_4 = 2(x_1 + x_0) + x_1 = 3x_1 + 2x_0$

Now we can start substituting in values for our algebra

$317,811 = 3x_1 + 2(46,368)$

We can now find a value for $x_1$

$\frac{317,811 - 92,736}{3} = x_1 = 75,025$

And now we have all the values for finding $n$

$n = x_1 + x_0 = 75,025 + 46,368 = 121,393$

So $F_{26} = 121,393$

Computer Science Solution in C++

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#include <iostream>
#include <vector>
using namespace std;
vector<long long> p;
int main()
{
   long long x;
   cin>>x;
   p.push_back(1);
   p.push_back(1);
   for(int i = 2; i < x; i++){
       p.push_back(p[i-1] + p[i-2]);
   }
   cout<<p[x-1];
} 

using the formula :Fn = $\frac{ϕ^n-(-1/ϕ)^n }{sqrt(5)}$ where ϕ is the golden ratio, this could help prove that , Fn + Fn+4 = 3Fn+2 ,which would help in the above question. Note that with this formula we can directly substitute the value of n and solve the question immediately no proof required.

Let F25 = x Then, F26 = 46368 + x. And, F27 = F25 + F26 = x + 46368 + x = 2x + 46368. It means that : F28 = 46368 + x + 2x + 46368. 317811 = 3x + 92736. 225075 = 3x. x = 75025.

So, F26 = 46368 + 75025. = 121393.

I'm kind of ignorant compared to some of the others who solved this. I got it right but it took a while and took trial and error. I would not recommend solving how I did.

First I hypothesized that f27 is 2/3 (66.6%) of f28. That every n is 2/3 of n+1. When checking this it was wrong. But when lowering the precentage from 66.6 to 61.8035(this number came from trial and error Untill taking 61.8035% of f28 5 times got close enough to f24). After that percentage was correct I just took it twice giving me 121, 393.397.

Not sure how coherent my answer was. I tried explaining the best I could. I know it was a bad way to solve but it was the only way I could figure out how and hey, it worked!

To make it even more easy,

Just visit HERE , get your number

Not very efficient, but... $F_{28} = F_{26} + F_{27}$

So, $317811 = F_{26} + F_{27}$

• We know that: $F_{26} = F_{24} + F_{25} = 46368 + F_{25}$
• We also know that: $F_{27} = F_{25} + F_{26} = F_{25} + (46368 + F_{25}) = 2 * F_{25} + 46368$

Now $317811 = (46368 + F_{25}) + (2 * F_{25} + 46368)$

Simplify: $317811 = 92736 + 3 * F_{25}$

Solve for $F_{25}$ : $F_{25} = 75025$

Thus, $F_{26} = 75025 + 46368 = \boxed{121393}$ .

F28 = F27 +F26

F26 =317811 - F27

F26 = 317811 - F26 -F25 ---- 1

F26= F25 +46368 ------- 2

adding 1 and 2 we get

3F26 =317811 +46368

F26=121393

If you're lazy, don't read everything I wrote, what really matters is the math notation. Most of these writings are just personal notations. By the way, I am not the most expert guy with math notation, so there may be a few ambiguities with the formal notation I love using.

It's a childish and rustic solution, I think, but it worked for me with what I knew. As a student, that's what I've been taught and that's what best suits for me. That's what makes me love math as well. Thank you, my teachers.

Well, knowing that:

$F_{n-1} + F_{n} = F_{n+1}$

and if $F_{n+1} = 5 \to (F_{n} = 3 \land F_{n-1} = 2) \because 2 + 3 = 5$

We also know that

$F_{24} = 46368$

$F_{28} = 317811$

and we wanna find $F_{26}$ . To find it, we need to know either it's antecedent or it's subsequent terms. We don't know them. If we wanna find $F_{26}$ we need to get rid of unknown terms. We still don't have enough information about these unknown terms, so I'll brainstorm some more among the things we know and can deduce by what we know.

$F_{24} + F_{25} = F_{26}$

$F_{25} + F_{26} = F_{27}$

$F_{26} + F_{27} = F_{28}$

That's enough. By that we can do some math black magic, and by that I mean isolate the unknown terms, so we can use them in substitution. It's important that we cannot isolate two terms from the same equation. Try doing this and using substitution and you'll get into an infinite loop or you'll get $\mathbb(R)$ . But it's ok, we have two equations for isolation and one for substitution. That's how I did it.

$F_{25} = F_{27} - F_{26}$ Here I thought: ok, I gotta find one unknown term, so I'll find $F_{25}$ , why not, right? So that's my starting point.

$F_{25} = F_{28} - F_{26} - F_{26}$ Substution

$F_{25} = F_{28} - 2F_{26}$ Simplification

$F_{26} - F_{24} = F_{28} - 2F_{26}$ Substitution

$-F_{24} = F_{28} - 3F_{26}$ Subtractive Property of Equality

$-3F_{26} + F_{28} = -F_{24}$ Symmetric Property of Equality

$-3F_{26} = -F_{28} - F_{24}$ * Subtractive Property of Equality

$F_{26} = \frac{F_{28} + F_{24}}{3}$ * Division Property of Equality

$F_{26} = \frac{317811 + 46368}{3}$ Substitution

$F_{26} = 121393$ Simplification

Of course there are some other ways of doing this same thing in a more neat way. It just happened everything flowed fine for me, but at the first time I didn't use this strategy I'm mentioning now, so that's why it's like this. I hope it's clear enough, but feel free to ask me to clarify even more, if it helps you or someone else understand. Some guys out there were finding some patterns I am very curious about and I'll certainly look into. Hope this helps.

Fibonacci series : 1 , 1 , 2 , 3 , 5 , 8 , 13 , ...... element = previous element + previous of previous element .... here if we take first five element as 1, 1, 2, 3, 5 ... so f1 = 1 & f5 = 5 , from that f3 = (1+5) / 3 = 2 . That's how every five pair works as in middle solution ; so the ans would be f26 = ( f24 + f28 ) / 3

Haha, I thought they said to not use a calculator. I had to use one.

F25=F24+F23

F26= F25+F24= 2F24+F23

F27= F26+F25 = 3F24+2F23

F28= F27+F26= 5F24+3F23

F28=317811. F24=46368

F28= 5F24+3F23

317811 =5(46368)+3F23

317811-5(46368)=3F23

85971/3=F23

F23=28657

F26=2F24+F23= 2(46368)+28657

F26=121393

Let's say F24 = A, F28 = B, and F25 = A + x. Then, F26 = 2A + x F27 = 3A + 2x F28 = 5A + 3x

So, B = 5A + 3x => B = 3* (2A + 3) - A => B = 3* F26 - A => F26 = (A+B)/3

But A = F24 = 46368 and B = F28 = 317811

So, F26 = (46368 +317811) / 3 = 121393 Answer