Question

Prove that (b^n)-1 is always divisible by (b-1) for all positive integers b and n.

Answer 1

b>1 as well.

Answer 2

1- show that it's true for n=1 ===> L.H.S = b^1 -1 = b-1 ===> it's divisib;e by (b-1).

Answer 3

2. assume it's true for n= k , i.e b^k -1 is divisible by b-1

Answer 4

3. show that it's true for n =k+1 ===> b^(k+1) is divisible by b-1, ===> b^(k+1) =b* b^k ===> from step 2. b^k is divisible by b-1 ===> b*b^kk is divisible by b-1 ===> b^(k+1) is divisible by b-1 ===> it's true for n>=1

Answer 5

this is what is called as proof using mathmatical induction.

Answer 6

hope that helps.

Answer 7

Nice proof.

Answer 8

great then.