Question

show that if n is even then
\(2^n-1 \) divides 3

Answer 1

let n =2k

Answer 2

\(\Large 4^k-1 = (4-1)(...)\)
done :)

Answer 3

haha nice

Answer 4

k belongs to integers, right ?

Answer 5

yeah lol 100%

Answer 6

mark why your cat is SO CUTE @Marki

Answer 7

ikr :)

Answer 8

anothe demostration can be made using the induction principle, namely, if I set n=2k, where k is a positive integer greater than zero, then I have:
\[k=1\rightarrow 2^{2k}-1=4-1=3\]
so our statement is true,
\[k=2\rightarrow 2^{2k}-1=16-1=15\]
so our statement is true again.
Now I suppose that our statement is true for a generic k, and I write that statement for k+1, namely:
\[2^{2(k+1)}-1=2^{2k}*4-4+4-1=(2^{2k}-1)*4+3\]
so our statement, by the induction principle is true again, and it is true for all k, or for all even n