Help me on proofs (induction) ?

Question

1. Show that 7^n -1 is divisible by 6, for all positive integers n.

Answer 1

there is an algebraic proof, but it would be difficult and i don't remeber it.

but there is another method to proof it, which i don't remeber its name.

for n=1, 7-1 = 6; for n=2, 49-8 mod 6 =0

we assume that for n, 7^n -1 mod 6 = 0 ( 7^n -1 is divisible by 6), then we prove that this for n+1 also holds.


7^(n+1) -1 = 7(7^n) -7 + 6 = 7(7^n -1) + 6

we assumed that 7^n -1 is divisibe by 6 so 7(7^n -1) is also divisible and also 7(7^n -1) + 6 is divisible

Answer 2

when n = 1: 7^1-1 = 6 is divisible by 6
suppose true until n, we have
7^(n+1) - 1 = 7*(7^n-1) + 6,
since 7^n-1 is divisible by 6, 7^(n+1)-1 is divisible by 6

Answer 3

t1=7-1=6 true
let it be true for some r,some positive integer
tr=7^r-1 true
tr+1=7^(r+1)-1
=7^r*7-1
add and subtract 7
7^r*7+7-7-1
(7^r-1)+6
since 7^r-1 and 6 are both divisible by 6 it is true for tr=1
so it is true for all positive integers of n

Answer 4

If it is true for two consecutive powers of n, it is true for all n. If n=1 7^1-1=6 If n=2 7^2-1=48 which is divisible by 6
therefore it is true

Answer 5

7 mod 6 = 1 thus 7^n mod 6 = 1 thus 7^n - 1 mod 6 = 0
voila.

Answer 6

7^2-7=7*6, since 7-1=6
so 7^2-1=(7^2-7)+6=(7+1)*6=8*6, divisible by 6.

(7^2-1)*7=7^3-7=8*6*7=56*6
so 7^3-1=(7^3-7)+6=(56+1)*6=57*6, divisible by 6.

(7^3-1)*7=7^4-7=57*6*7
so 7^4-1=(7^4-7)+6=(399+1)*6=400*6, divisible by 6.

You get it?!