prove by induction that 4^n-1 is divisible by three for all posotive integers of n
any help would be appriciated thanks.
prove $\displaystyle 4^n - 1$ is divisible by 3
true for $\displaystyle n = 1$
assume true for $\displaystyle n$, show true for $\displaystyle (n+1)$ ...
since $\displaystyle 4^n - 1$ is divisible by 3, then $\displaystyle 4^n - 1 = 3k$ , $\displaystyle k$ a positive integer.
$\displaystyle 4^n - 1 = 3k$
$\displaystyle 4(4^n - 1) = 4(3k)$
$\displaystyle 4^{n+1} - 4 = 12k$
$\displaystyle 4^{n+1} - 1 - 3 = 12k$
$\displaystyle 4^{n+1} - 1 = 12k + 3$
$\displaystyle 4^{n+1} - 1 = 3(4k + 1)$
since $\displaystyle (4k+1)$ is also a positive integer, $\displaystyle 4^{n+1} - 1$ is a multiple of 3, and therefore, is divisible by 3.
Hello, hmmmm!
Here's another way . . .
Prove by induction that $\displaystyle 3|(4^n-1)$ for all positive integers $\displaystyle n.$
Verify $\displaystyle S(1)\!:\;\;4^1-1 \:=\:3$ . . . true
Assume $\displaystyle S(k)\!:\;\;4^k-1 \:=\:3a\,\text{ for some integer }a.$
Add $\displaystyle 3\!\cdot\!4^k$ to both sides: .$\displaystyle {\color{blue}3\!\cdot\!4^k} + 4^k - 1 \;=\;3a + {\color{blue}3\!\cdot\!4^k}$
Factor: .$\displaystyle (3 + 1)4^k - 1 \;=\;3\left(a + 4^k\right)$
. . . . . . . . $\displaystyle 4\!\cdot4^k - 1 \;=\;3\left(a + 4^k\right)$
. . . . . . . . $\displaystyle 4^{k+1}-1 \;=\;\underbrace{3\left(a + r^k\right)}_{\text{a multiple of 3}}$
We have proved $\displaystyle S(k+1)$ . . . The inductive proof is complete.