Answer Happy • for n in N, we have Fn = 2^2^n + 1 and m in N with m>n a) Proove that 2^2^n -1(fn), and that (2^2^n)^(2^(m-n)) = fm -1

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for n in N, we have Fn = 2^2^n + 1 and m in N with m>n a) Proove that 2^2^n -1(fn), and that (2^2^n)^(2^(m-n)) = fm -1

Posted: Mon Apr 11, 2022 6:05 am

by answerhappygod

for n in N, we have Fn = 2^2^n + 1 and m in N with m>n
a) Proove that 2^2^n

: For N In N We Have Fn 2 2 N 1 And M In N With M N A Proove That 2 2 N 1 Fn And That 2 2 N 2 M N Fm 1 1 (1.3 KiB) Viewed 40 times

-1(fn), and that
(2^2^n)^(2^(m-n)) = fm -1
b) show that fm

: For N In N We Have Fn 2 2 N 1 And M In N With M N A Proove That 2 2 N 1 Fn And That 2 2 N 2 M N Fm 1 2 (1.3 KiB) Viewed 40 times

2(fn)
c) now, we want proove that
fm

: For N In N We Have Fn 2 2 N 1 And M In N With M N A Proove That 2 2 N 1 Fn And That 2 2 N 2 M N Fm 1 3 (1.3 KiB) Viewed 40 times

fn =1
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