prove by induction 3^n -1 is even?

P(n): 3^n - 1 is even
Where n is a positive integer.

When n = 1 :
3*1 - 1 = 3 - 1 = 2
As 2 is a positive integer, P(1) is true.

Assume that P(k) is true, i.e.
3^k - 1 = 2Q
where Q is a positive integer.

When n = k + 1 :
3^(k + 1) - 1
= 3×3^k - 1
= 3×(3^k - 1) + 3×1 - 1
= 3×2Q + 2
= 2 × (3Q + 1)
= an even number when P(k) is true.
Hence, P(k + 1) is true when P(k) is true.

According to the principle of mathematical, 3^n - 1 is even for all positive integers n.

Base case: n =1 so 3¹ - 1 = 3 - 1 = 2 which is even so the expression is true for n=1.
Assume: 3ᴷ-1 is even.
Now consider 3ᴷ⁺¹-1.
3ᴷ⁺¹-1 = 3(3ᴷ)-1 = 3(3ᴷ)-3+2 = 3(3ᴷ-1)+2
By assumption 3ᴷ-1 is even so 3(3ᴷ-1) is even.
2 is even.
An even number + an even number is even so 3(3ᴷ-1)+2 is even.
Thus if the expression is true for the Kth case, it is true for the K+1 case.
The expression is true for n=1, so it is true by induction for all positive integers.

The usual proof by induction is to show that the expression is true for n=1 (or some other base value), then show that if it is true for any arbitrary number n, it is also true for n+1.

For n = 1, 3ⁿ - 1 = 3¹ - 1 = 3 - 1 = 2.
For arbitrary n (where n ≥ 2), 3ⁿ = 3ⁿ⁻¹·3¹. Any power of 3 is an odd number (since the product of two or more odd numbers is odd), so 3ⁿ⁻¹ is odd, so 3ⁿ⁻¹·3¹ is odd, so 3ⁿ⁻¹·3¹ - 1 is even (because 1 less than any odd number is an even number).
For n+1, 3ⁿ⁺¹ = 3ⁿ·3¹, which is again an odd number, so 3ⁿ⁺¹ - 1 is also even.

Question is dodgy !

Is it [ 3^n ] - 1 or is it 3^(n - 1) ????

This subject is not about guessing.