<s>[INST] 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 $n\ge0$
 [/INST] Direct Proof:
$3^n$ has no factor of $2$, so it is odd. $3^n+1$ is one greater than an odd number.

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.
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