Verified answer

Can not be proved because the statement is wrong!

F[1] = 1, F[2] = 2 , F[3] =3, F[4] = 5, F[5] = 8 , F[6] = 13.

But for n=3, Σ ⁿ _(i=1) F[2i-1] = F[1] + F[3] + F[5] = 12 , but F[6] = 13.

The correct statement is: Σ (i=1 to n) F[2i-1] = F[2n] - 1.

Can be easily proved by induction:

Checked for n = 1: as F[1] = 1 = F[2] -1.

Assume true for n-1: F[1] + F[3] +....F[2n-3] = F[2n-2] -1. And based on this:

(F[1] + F[3] +....F[2n-3]) + F[2n-1] = ( F[2n-2] -1) + F[2n-1] = F[2n] -1 as required!

I think Fibonacci sequence is f1 = 1, f2 = 1 and f3 = f1 + f2 = 3 and so on

Notation: i'm employing _ as a subscript indicator to designate words interior the sequence. a) (fn,fn+a million)= a million i assume the () subsequently is meant to point the terrific elementary ingredient. assume that for some fee of n, (f_n, f_(n+a million)) = ok > a million this is, some pair of successive words interior the sequence have a elementary ingredient extra suitable than a million. Then because ok divides the two f_n and f_(n+a million), it needless to say additionally divides f_(n-a million) = f_(n+a million) - f_n and it additionally divides f_(n+2) = f_n + f_(n+a million) and the comparable reasoning enables us to amplify divisibility by ability of alright to all words interior the sequence. yet it seems that the 1st 2 words are no longer divisible by ability of ok if ok>a million, so our supposition that 2 successive words have a elementary ingredient extra suitable than a million ends up in a contradiction and would desire to 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 assume this formulation holds for some n=ok; this 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 the comparable formulation for n=ok+a million. because we can relatively show that the formulation holds for n=a million: a million^2 = a million*2 + (-a million)^a million the formulation holds for all integers n>a million by ability of mathematical induction.