N = 10, = 100, = 1000 in a range

Question

I have the the sum from (i = 1) to N is 1 + 2 + 3 + 4 ... + N

I found this program to calculate the sum for i in range(1,100)

num1, num2 = 1, 100
sum = int((num2*(num2+1)/2) - (num1*(num1+1)/2) + num1)
print(sum)

This works, but what if I want to know N = 10, or N = 100?

Jiho Kim · Answer 1 · 2022-11-08T23:16:21.513

1

Here is a better way to do that:

def sum(n):
    return int(n * (n + 1) // 2)

print(sum(10))
print(sum(100))

edited Nov 08 '22 at 23:16

answered Nov 08 '22 at 22:51

Jiho Kim

158
8

1

You could just `return int(...)` rather than wrap it into `int()` everytime you call the function. – Jimpsoni Nov 08 '22 at 22:53
1

@Jimpsoni Thanks for pointing that out! Edited accordingly. – Jiho Kim Nov 08 '22 at 22:55

score 0 · Answer 2 · answered Nov 08 '22 at 22:54

This program works by pairing the numbers together from the end to the beginning. For example, to find the sum of the list 1, 2, 3, 4, 5, 6, you can pair 1 and 6 to make 7, 2 and 5 to make 7, and 3 and 4 to make 7. See a pattern? (hint: the pairs always add up to the sum of the first and the last).

This program also includes an adjustment for lists with an odd number of elements. For example, in the list 1, 2, 3, one needs to add the additional middle number that cannot be in a pair. 1 and 3 can pair to 4, but you'll need to add the unpaired 2 to the sum.

Now that you understand it, all you need to do is extract it into a function. Like this:

def consecutiveSum(num2):
    num1 = 1
    return int((num2*(num2+1)/2) - (num1*(num1+1)/2) + num1)


print(consecutiveSum(10))  # => 55
print(consecutiveSum(100))  # => 5050

Jiho Kim's answer also points out that this can be optimized a bit. See his answer too.

N = 10, = 100, = 1000 in a range

2 Answers2