Shortest Python 3 for sum of arithmetic progression

Question

Input: from STDIN number of vertices in Graph \$2 \leq N \leq 100\$.

Rules: [Code size] = max ([code length without spaces, tabs and newlines], [total code length divided by 4])

Math formulation: In the graph of N vertices, between each pair of vertices can be 3 road states:

• there is no road
• there is a road from A to B
• there is a road from B to A

Find the number of different graphs on given vertices.

We can apply next formula (number of different road states in pow of pairs number): $$\huge3 ^ {\frac{n(n - 1)}{2}}$$.

My Python 3 37 bytes solution here:

n = int(input())
print(3 ** ((n ** 2 - n) // 2))

I know that exists 34 bytes solution. Then I started to think about 1 liner, and find next formula for sum of arithmetic progression, which use N only once: $$\frac{(2n - 1) ^ 2}{8} - \frac18$$

Unfortunately the code only increased:

x = 2 * int(input()) - 1
print(3 ** ((x ** 2 - 1) // 8))

Answer 1

Python 3, 34 bytes

print(3**sum(range(int(input()))))

Try it online!

In Python 2 this is just print 3**sum(range(input())) for 28.

Answer 2

Python 3, 34 bytes (not counting the newline)

Uses the conventional formula for the sum of the arithmetic progression (\$\frac{n(n-1)}2\$) and the trick that n-1 = ~-n (but with higher operator precedence). This way, one pair of parentheses can be omitted.

n=int(input())
print(3**(n*~-n//2))

Try it online!