Sum of first N natural numbers in Python

The natural numbers are 1, 2, 3 and so on. This program adds the first n of them, first with a loop and then with a formula that needs no loop at all.

The program

n = int(input("Add the numbers from 1 to: "))

total = 0
i = 1
while i <= n:
    total += i
    i += 1

print(f"Sum of 1 to {n} = {total}")

Typed in this example: 100. Press Run to type your own.

Add the numbers from 1 to: 100
Sum of 1 to 100 = 5050

How it works

• total starts at 0. Each pass of the loop adds the current i to it, then moves i on by one.
• The loop stops when i becomes larger than n.
• If the user types 0, the loop body never runs and the sum is 0, which is the correct answer.

Two shorter ways

sum() adds up everything in a range. The formula n * (n + 1) // 2 gives the same answer with one multiplication, however large n is.

n = 100

print(sum(range(1, n + 1)))
print(n * (n + 1) // 2)

5050
5050

The formula works because you can pair the numbers from both ends: 1 + 100, 2 + 99, 3 + 98, and so on. Each pair adds up to n + 1, and there are n / 2 pairs. // is whole-number division, so the result is an int and not 5050.0.

Related programs: Sum of a series.