Proof by induction that $3^n - 1$ is an even number

Question

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

Answer 1

Hint: try substituting back in $3^k = 2x+1$

Answer 2

Hint: Start by showing that $3^n$ is odd. What's an odd minus an odd?

Answer 3

Hint : $3^{k+1}-1=3(3^{k})-1=3(3^{k}-1)+3-1=3(3^{k}-1)+2$

Answer 4

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.

Answer 5

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 $\,1^n\equiv 1.$

Answer 6

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 $n > 0$.

Answer 7

Note $3^1-1=2$ is even. Then:

$$\begin{align} \text{Given } &3^k-1 \text{ is even} \\ \implies 3(3^k-1) &= 3^{k+1}-3 \text{ is even} \\ \implies 3^{k+1}-3+2 &= 3^{k+1}-1\text{ is even} \\ &\square \end{align}$$

Answer 8

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.

Answer 9

$3^{k+1}$ 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.