The Fibonacci sequence is defined as F1 = 1, F2 = 1, Fn+1 = Fn + Fn-1, n>=2. Determine the value of F26.?

Question

Please show your working in the solution and explain how you arrive at that answer. Thank you!!!

Answer 1

F3= F2+F1 = 1+1=2
F4=F3+F2=2+1 =3
F5=3+2=5
F6=8
F7=13
F8=21
F9=34
F10=55
F11=89
F12=144
F13=233
F14=377
F15=610
F16=987
F17=1597
F18=2584
F19=4181
F20=6765
F21=10946
F22=17711
F23=28657
F24=46368
F25=75025

F26=121393, were my calculation right.

Answer 2

There's another way to do this, that doesn't require
writing down 26 terms of the Fibonacci sequence.
Consider the associated Lucas sequence
L_1 = 1, L_2 = 3, L_n+1 = L_n + L_n-1.
Write the first 13 terms of each sequence
L_n F_n
1 1
3 1
4 2
7 3
11 5
18 8
29 13
47 21
76 34
123 55
199 89
322 144
521 233

If you look carefully, you will notice the
following identity:
F_2n = F_n*L_n,
which can be proved either by induction
or by using Binet's formula(see below).
So F_26 = F_13*L_13 = 521*233 = 121393.

Binet's formula:
Let α = (1 + √5)/2, β = (1 - √5)/2
Then
L_n = α^n + β^n and F_ n = (α^n - β^n)/√5.
Now note that if you multiply L_n and F_n you
get the formula for F_2n.
Hope that helps!

Answer 3

Notation: i'm utilising _ as a subscript indicator to designate words interior the sequence. a) (fn,fn+a million)= a million i assume the () thus is meant to point the most excellent elementary element. think that for some fee of n, (f_n, f_(n+a million)) = ok > a million it truly is, some pair of successive words interior the sequence have a elementary element more advantageous than a million. Then on the grounds that ok divides both f_n and f_(n+a million), it obviously also divides f_(n-a million) = f_(n+a million) - f_n and it also divides f_(n+2) = f_n + f_(n+a million) and a similar reasoning enables us to enlarge divisibility with techniques from okay to all words interior the sequence. yet it seems that the first 2 words are literally not divisible with techniques from ok if ok>a million, so our supposition that 2 successive words have a elementary element more advantageous than a million finally ends up in a contradiction and must be fake. hence, for any n, (f_n, f_(n+a million)) = a million b) (f_(n+a million))^2= f_n*f_(n+2) + (-a million)^n think this formula holds for some n=ok; it truly is, [f_(ok+a million)]^2 = f_k f_(ok+2) + (-a million)^ok Then (f_(ok+2))^2 = [f_(ok+a million) + f_k] [f_(ok+3) - f_(ok+a million)] = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+3) - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k [f_(ok+2) + f_(ok+a million)] - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+2) + f_k f_(ok+a million) - f_k f_(ok+a million) = f_(ok+a million) f_(ok+3) - [f_(ok+a million)]^2 + f_k f_(ok+2) = f_(ok+a million) f_(ok+3) - [f_k f_(ok+2) + (-a million)^ok] + f_k f_(ok+2) = f_(ok+a million) f_(ok+3) - f_k f_(ok+2) + (-a million)^ok + f_k f_(ok+2) - (-a million)^ok = f_(ok+a million) f_(ok+3) + (-a million)(-a million)^ok = f_(ok+a million) f_(ok+3) + (-a million)^(ok+a million) this is an identical formula for n=ok+a million. on the grounds that we may be able to somewhat show that the formula holds for n=a million: a million^2 = a million*2 + (-a million)^a million the formula holds for all integers n>a million with techniques from mathematical induction.

Answer 4

F(26) = 121393

Just keep adding the previous two numbers to get the next

Answer 5

.. holy hell.