Mathematical Induction: $3^n +1$ is even for all values of positive integers 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 $n\ge0$

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

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
