Proof by induction that $3^n - 1$ is an even number How to demonstrate that  $3^n - 1$ is an even number using the principle of induction? 
I tried taking that $3^k - 1$ is an even number and as a thesis I must demonstrate that $3^{k+1} - 1$ is an even number, but I can't make a logical argument. 
So I think it's wrong the assumption... 
Thanks
 A: Hint: Start by showing that $3^n$ is odd. What's an odd minus an odd?
A: Hint : $3^{k+1}-1=3(3^{k})-1=3(3^{k}-1)+3-1=3(3^{k}-1)+2$
A: Hint: try substituting back in $3^k = 2x+1$
A: First,  $ 3^0 - 1 = 0 $ which is even so $3^n - 1 $ is even for $ n = 0 $
Suppose that 
$3^k - 1 $ is even. We need to show that $3^{k+1}- 1 $ is even. Well, 
$ 3^{k+1}-1 = 3(3^{k}) -1 = 3(3^k - 1) + 2 $  
By our assumption, $ 3^k-1 $ is even so $ 3^{k} - 1 = 2m $, for some integer $m$
Thus, $ 3(3^k - 1) + 2 = 3(2m) + 2 = 2(3m + 1) $ Note that, $3m + 1 \in Z $ since addition and multiplication are closed in $Z$. Thus, $2(3m + 1) = 3^{k+1} -1 $ is even which is want we need to show.
Therefore, $3^n -1$ is even for all $n > 0$.
A: Hint $\ $ Let $\,a = 2\,$ below (= first term of binomial expansion).
$\ \ \ \begin{align} (1+ a)^n\, \ \  =&\,\ \ 1 + ak\qquad\qquad\quad {\rm i.e.}\ \ P(n)\\[1pt]
\Rightarrow\ (1+a)^{\color{#c00}{n+1}}\! =&\  (1+ak)(1 + a)\\[2pt]  
 =&\,\ \ 1+ a\,(\underbrace{k\!+\!1\!+\!ka\!}_{\large k'})\ \ \ {\rm i.e.}\ \ P(\color{#c00}{n\!+\!1})\\  
  \end{align}$
Remark $\ $ If you know modular arithmetic (congruences) then you can view it as a special case of the Congruence Power Rule, i.e. $\, x\equiv 1\,\Rightarrow\, x^n\equiv 1^n\equiv 1\pmod{\!a},\,$ where the induction is conceptually clearer: $ $ the powering of a congruence, and the trivial induction $\,1^n\equiv 1.$
A: Let $3^k-1=2m$ for some $m\in\Bbb Z$.
$$3^{k+1}-1=3(3^k-1)+2=2(3m+1)$$  
I definitely wouldn't use induction here though.
A: Note $3^1-1=2$ is even. Then:
$$\begin{align}
\text{Given } &3^k-1 \text{ is even} \\
\implies 3(3^k-1) &= 3^{k+1}-3 \text{ is even} \\
\implies 3^{k+1}-3+2 &= 3^{k+1}-1\text{ is even} \\
&\square
\end{align}$$ 
A: We assume that $3^k-1$ is even. This is the assumption of induction. We want to show that $3^{k+1}-1$ is even. We can rewrite this as $3 \cdot 3^k - 1$. Now calculate the difference between the two numbers:
$$(3\cdot 3^{k} - 1) - (3^k-1) = 3\cdot 3^k - 1 - 3^k + 1 = 3\cdot 3^k - 3^k$$ Now we can factor out $3^k$ so we get: $3^k(3-1)= 3^k2$, which is an even number. If you add an even number to an even number you always end up with an even number. Therefore the induction step is now complete.
A: $3^{k+1}$ and $3^k$ have the same parity as$$3^{k+1}-3^k=2\cdot3^k,$$
and so do $3^{k+1}-1$ and $3^k-1$.
The base case is $3^0-1=0$, even.
