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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

N. F. Taussig
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ColdFish
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9 Answers9

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

syusim
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    ...except that is not induction. – Joffan Jun 09 '15 at 20:58
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    How so? $3^0-1 = 0$ is even. For $n \geq 1, 3^{n-1}-1$ is even. Then $3^{n-1}$ is odd. Then $3^n$ is odd. Then $3^n-1$ is even. – Brian Tung Jun 09 '15 at 21:08
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    @BrianTung The problem is so simple that any random observation could be worked into induction; nevertheless the answer above does not do that. – Joffan Jun 09 '15 at 21:38
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    It doesn't, but I would guess that the OP understands how to work things into induction, and is mostly having difficulty framing the premise of the induction in a convenient form. This answer nudges the OP toward one such convenient form. – Brian Tung Jun 09 '15 at 22:08
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Hint : $3^{k+1}-1=3(3^{k})-1=3(3^{k}-1)+3-1=3(3^{k}-1)+2$

mich95
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Hint: try substituting back in $3^k = 2x+1$

nosyarg
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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 $n > 0$.

jayd0gw
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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 $\,1^n\equiv 1.$

Community
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Bill Dubuque
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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.

user26486
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Note $3^1-1=2$ is even. Then:

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

Joffan
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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.

mathreadler
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$3^{k+1}$ 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.