Sum of the integers from 1 to n

Question

I'm trying to write a program to add up the numbers from 1 to n. I've managed to get it to print the numbers several times but not to add them all. It keeps on just adding two of the numbers.

My 1st attempt is:

def problem1_3(n):
    my_sum = 0
    while my_sum <= n:
        my_sum = my_sum + (my_sum + 1)
    print() 
    print(my_sum)

How can I fix this problem?

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^{For the recursive version of this question, see Recursive function to calculate sum of 1 to n?}

Answer 1

There is no need for a loop at all. You can use the triangular number formula:

n = int(input())
print(n * (n + 1) // 2)

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A note about the division (//) (in Python 3): As you might know, there are two types of division operators in Python. In short, / will give a float result and // will give an int. In this case, we could use both operators, the only difference will be the returned type, but not the value. Since multiplying an odd with an even always gives an even number, dividing that by 2 will always be a whole number. In other words - n*(n+1) // 2 == n*(n+1) / 2 (but one would be x and the other x.0, respectively).

Answer 2

You can do it with one line, where you sum the range of numbers from 0 to n (the end is exclusive):

def problem1_3(n):
    return sum(range(n+1))

Answer 3

You need 2 different variables in your code -- a variable where you can store the sum as you iterate through the values and add them (my_sum in my code), and another variable (i in my code) to iterate over the numbers from 0 to n.

def problem1_3(n):
    my_sum = 0
    i=0
    #replace this pass (a do-nothing) statement with your code
    while i <= n:
        my_sum = my_sum + i
        print(my_sum)
        i+=1
    return my_sum

You are using the my_sum variable in your code to both store the sum and iterate through the numbers.

Answer 4

You could use numpy to sum the numbers in the arange of 1 to n+1 (exclusive):

import numpy as np 

np.sum(np.arange(1, n+1))

Answer 5

The sum of numbers from 1 to n will be greater than n. For example, the sum of numbers from 1 to 5 is 15 which is obviously greater than 5. Your while loop terminates prematurely. You need to maintain a separate counter for the loop.

Answer 6

Real programmers use recursion (and hopes for a not too big n since there is no tail call optimization in Python):

def problem1_3(n):
    return n + problem1_3(n-1) if n > 1 else 1

Answer 7

so it will be more optimal

def f(a):
    return (a + 1) * (abs(a) + 2 * (a <= 0)) // 2

Answer 8

This while loop works:

def main():
    n = int(input('Enter a number:'))              
    while n <= 1:
        n = int(input('Please input a new number'))

    total = 0
    my_sum = 0
    while (n-1) >= total:
        total = total + 1
        my_sum += total
        print(my_sum)
        
    return 

main()

Answer 9

In this case where you iterate over a known sequence, it is better to use a for loop and add the numbers as you go:

def summation(num): 
    sumX = 0
    for i in range(1, num + 1):
        sumX = sumX + i
    return sumX

summation(3)  # You place the num you need here

Answer 10

n = input("Enter Number to calculate sum")
n = int (n)
#average = 0.
#sum = 0
sum = 0.
for num in range(0,n+1,1):
    sum = sum+num
print("SUM of first ", n, "numbers is: ", sum )

# Print sum of numbers from 1 to N inclusive

def sum2N(N):
    r = 0
    for i in range(N+1):
        #for i in range(0,N+1,1):
        #r+=i
        r=r+i
    return(r)

print(sum2N(10))

Answer 11

How about you try it using a "While Loop":

def problem1_3(n):
my_sum = 0
while my_sum <= n:
    print(my_sum,end=" ")  # end = " " keeps from starting a new line
    my_sum = my_sum + 1
print(my_sum)