1. Mathematical induction

Prove (7^n)-1 is divisible by 6,n>0

2. Assume $\displaystyle 7^{n}-1=6p$

Then show that $\displaystyle 7^{n+1}-1=6k$

$\displaystyle 7^{n}-1=6k$

$\displaystyle 7\cdot (7^{n}-1)=7\cdot 6k$

$\displaystyle 7^{n+1}-7=6\cdot 7k$

$\displaystyle 7^{n+1}-1=6\cdot 7k+6$

$\displaystyle 7^{n+1}-1=6(7k+1)$...QED

3. Hello, Knigh17!

A variation of galactus' solution . . .

Prove $\displaystyle 7^n-1$ is divisible by 6, $\displaystyle n>0$

Verify $\displaystyle S(1)\!:\;\;7^1 - 1 \:=\:6$ . . . true.

Assume $\displaystyle S(k)\!:\;\;7^k - 1 \:=\:6a\;\;\text{ for some integer }a$

Add $\displaystyle 6\!\cdot\!7^k$ to both sides: .$\displaystyle 7^k + {\color{blue}6\!\cdot\!7^k} - 1 \;=\;6a + {\color{blue}6\!\cdot\!7^k}$

We have: .$\displaystyle (1 + 6)\!\cdot\!7^k - 1 \;=\; 6a + 6\!\cdot\!7^k$

. . . . . . . . . . . $\displaystyle 7\!\cdot\!7^k - 1 \;=\;6a + 6\!\cdot\!7^k$

. . . . . . . . . . . $\displaystyle 7^{k+1} - 1 \;=\;\underbrace{6(a + 7^k)}_{\text{a multiple of 6}}\quad\hdots\;QED$

4. Thanks

Thanks a lot! now i know how this is solved.