Sum of the integers from 1 to n

Question

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

My 1st attempt is:

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

How can I fix this problem?

Answer 1

I don't understand why everyone keeps making everything complex. Here is my simple solution

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

Answer 2

You can do it with one line, where you create a list of integers from 0 to n and sums all the elements with sum function

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

Answer 3

That's where I use "math" to solve problems like this. There is a formula for solving this problem: n * (n+1) / 2.

The code I would write:

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

Answer 4

This one line do the job :

sum(range(1, n+1))

Answer 5

You could use numpy.

import numpy as np 

Your function should return

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

https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.sum.html https://docs.scipy.org/doc/numpy-1.15.0/reference/generated/numpy.arange.html

Answer 6

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 7

Actually, I have tried a lot of that type of Programs In Jupyter Notebook you may use these:

n = int(input("Enter the Number: "))
print(n * (n + 1) // 2)

And Also You may try this code:

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

And Also You may try it

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

Answer 8

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 9

so it will be more optimal

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

Answer 10

print("Sum =",sum(range(int(input("Enter a number\n"))+1)))

Answer 11

THIS WHILE LOOP ACTUALLY 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 12

These lines worked for me in Python

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

As well as these:

def summation(num):
    return sum(range(1, num+1))

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

Answer 13

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) 

Answer 14

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))