Mathematical Induction: $3^n +1$ is even for all values of positive integers

Question

I am not sure how to go about this "proof by induction" problem. below is my attempt.

Base Case: $n = 0$,substituting the value of $n$ to the equation $3^n+1$ $$= 3^0 + 1$$ $$= 1 + 1 = 2 $$ Thus the equation holds true for initial value of $n$ i.e $0$

Induction Hypothesis: Suppose the equation holds true for all the values of $n$: $1,2,3....k$ therefore, $3^k + 1$ results even.

Induction Step: $n = k$ holds true,
to prove: $3^{k+1}+ 1$ for $n = k+1$

LHS: $$3^{k+1}+ 1 = (3^k \cdot 3) + 1 = (3^k \cdot (2 + 1)) + 1 = 2 \cdot 3^k + 3^k + 1$$

The above solution results even, because since multiplying any integer with $2$ gives even integer, and from the Induction Hypothesis $3^k+1$ is even.

Hence $3^{k+1} + 1$ is even, thus $3^{n+1}$ is even for all values of $n\ge0$

Answer 1

Direct Proof:

$3^n$ has no factor of $2$, so it is odd. $3^n+1$ is one greater than an odd number.

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Inductive Proof:

$3^0+1=2$ is even.

Suppose $3^n+1$ is even, then $$ 3^{n+1}+1=3\left(3^n+1\right)-2 $$ is an even number minus an even number, hence even.

Answer 2

Proof I

$$3^n + 1 \implies (2x + 1)+1 \implies 2x+2$$

Case: $n = 0$ $$ (2\cdot0 + 2) = 2 $$ Case: $n = n$ $$(2n + 2) = 2(n+1) $$ Case: $n = n+1$ $$(2(n+1) + 2) = 2n+4 = 2(n+2)$$ $$\Box$$

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Proof II

$$(3^n + 1)\text{ mod }2 \implies (3^n + 2 -1) \text{ mod }2 \implies (3^n -1)\text{ mod }2 $$

Case: $n = 0$ $$3^0 \equiv 1 (\text{ mod }2)\implies 3^0 -1 \equiv 0 (\text{ mod }2)$$

Case: $n = n$ $$3^n \equiv 1 (\text{ mod }2)\implies 3^n -1 \equiv 0 (\text{ mod }2)$$ Case: $n = n+1$ $$3^{n+1} \equiv 1 (\text{ mod }2)\implies 3^{n+1} -1 \equiv 0 (\text{ mod }2)$$ $$\Box$$

Case of the Congruence Power Rule

Answer 3

An easier answer is to note that:

$3^n - 1 \equiv 1^n - 1 \equiv 0 (mod 2)$.Thus, we conclude that $2|(3^n -1)$. Hence, $3^n - 1$ is always even.