prove by induction that 4^n-1 is divisible by three for all posotive integers of n

any help would be appriciated thanks.

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- Jan 28th 2009, 07:38 AMhmmmmproof by induction
prove by induction that 4^n-1 is divisible by three for all posotive integers of n

any help would be appriciated thanks. - Jan 28th 2009, 07:51 AMrunning-gag
Hi

Initialization of induction is easy.

Let's suppose that for a given n $\displaystyle 4^{n}-1$ is divisible by 3.

You need to demonstrate that $\displaystyle 4^{n+1}-1$ is divisible by 3.

Hint : $\displaystyle 4^{n+1}-1 = 4(4^{n}-1)+3$. - Jan 28th 2009, 08:01 AMskeeter
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. - Jan 28th 2009, 08:11 AMSoroban
Hello, hmmmm!

Here's another way . . .

Quote:

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.