## Sum of the integers from 1 to n

sum of numbers from 1 to n calculator

find the sum of integers from 1 to 100

n(n+1)/2

sum of numbers from 1 to 99

sum of all numbers -1/12 proof

sum of 1/2^n

sum of numbers from 1 to 100 in c

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?

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.

**Sum of the First n Natural Numbers,** In other words, is there a way to quickly calculate the sum of all the integers from 1 up to any other number—which we'll call “n”—that your friends The Sum of Positive Integers Calculator is used to calculate the sum of first n numbers or the sum of consecutive positive integers from n 1 to n 2. Sum of Consecutive Positive Integers Formula The sum of the first n numbers is equal to: n(n + 1) / 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))

**How to Quickly Add the Integers From 1 to n?,** Question: what is the sum of the first 100 whole numbers?? how am i in the late 1700's, Gauss was asked to find the sum of the numbers from 1 to 100. S is the sum of the series and n is the number of terms in the series, in this case, 100. a = 1,n = 100 is famously said to have been solved by Gauss as a young schoolboy: given the tedious task of adding the first. 100 positive integers, Gauss quickly used a formula to calculate the sum of. The formulas for the first few values of. a are as follows: ∑ k = 1 n k = n ( n + 1) 2 ∑ k = 1 n k 2 = n ( n + 1) ( 2 n + 1) 6 ∑ k = 1 n

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

**1 + 2 + 3 + 4 + ⋯,** a young schoolboy: given the tedious task of adding the first 100 100 100 positive integers, Gauss quickly used a formula to calculate the sum of 5050. 5050. 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.

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.

**What is the sum of the first 100 whole numbers?,** Given a number n, find sum of digits in all numbers from 1 to n. Examples: Input: n = 5 Output: Sum of digits in numbers from 1 to 5 = 15 Input: n = 12 Output: Sum Recap: Adding the Integers From 1 to 100. Before we figure out how to add up all the integers from 1 to n, let’s recap how to add up all the integers from 1 to 100.The key to this is our friend the associative property of addition which says that you are free to add together a group of numbers in any order you like.

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

**Sum of n, n², or n³,** As the top row increases, the bottom row decreases, so the sum stays the same. Because 1 is paired with 10 (our n), we can say that each column has (n+1). And The base case shown by the applet is n=1, although on the proof pages the base case is n=0; this is just because there is nothing to show when n=0. Assuming the result for n means we know how to sum half of an n x (n+1) rectangle having rows with 1, 2, , n red dots, respectively.

**Compute sum of digits in all numbers from 1 to n,** The sum of integers from 1 to n (inclusive) is n(n+1)/2. What is the sum from 50 to 100 (inclusive) ? --== Message from the GMAT Club Team ==-- n(n + 1)/2 = Sum of Integers In this case, n=100, thus you get your answer by entering 100 in the formula like this: 100(100 + 1)/2 = 5,050 Sum of Integers from 1 to 101

**Sum of "n" Consecutive Integers - Simple Proof,** Sum of integers from 1 to n = [n*(n + 1)]/2. Using this formula sum of integers from 1 to 100 = [100*(100 + 1)]/2 = (100*101)/2 = 5050. You can see how this formula works by taking a simple

**Techniques for Adding the Numbers 1 to 100 – BetterExplained,** Enter a positive integer: 50 Sum = 1275. This program assumes that user always enters positive number. If user enters negative number, Sum = 0 is displayed and program is terminated. This program can also be done using recursion. Check out this article for calculating sum of natural numbers using recursion.

##### Comments

`my_sum = my_sum + (my_sum + 1)`

doubles your previous value and adds`1`

to it. Why not`my_sum += 1`

(which is equivalent to`my_sum = my_sum + 1`

). Outside of that, your indentation is off and you don't show how you call the function.`while my_sum <= n`

This condition doesn't look correct. Did you mean to use a counter there? Consider 1 2 3 4. This will get out of the loop at 3 because the sum (1+2+3) will be larger than 4.- (1) Here's a much faster solution:
`return n*(n+1)/2`

. (2) Please fix your code indentation. (3)`my_sum`

is the sum of numbers. You want to loop through the numbers themselves. So you'll need another variable in there - With Python3, range produces a generator not a list, I thought it had to be explicitly converted to a list for sum, but without it it also works. Thanks for the remark.
- No worries. Most standard functions & methods that accept a list will also accept a generator, and it's generally more efficient to feed them a generator if you can. So (for example) you can pass a generator to
`max`

. One big exception is the`str.join`

method. It*can*accept a generator, but it needs to scan the sequence of strings twice: the 1st time it calculates the total length of the final destination string, the 2nd time it copies the data. So if you pass it a generator it has to build a list before it can process the data. The same goes for the`bytes.join`

method. - (cont) Similarly,
`sorted`

can accept a generator, but it has to construct a list, which it then calls the`list.sort`

method on, so there's no advantage in using`sorted`

on a generator, rather than`.sort`

on a list, apart from the more compact syntax. - And, yes, as it is currently written, this function wouldn't be able to use TCO, anyway...
- Welcome to Stack Overflow! Please don't answer just with source code. Try to provide a nice description about how your solution works. See: How do I write a good answer?. Thanks
- Welcome to Stack Overflow! Thank you for the code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by describing why this is a good solution to the problem, and would make it more useful to future readers with other similar questions. Please edit your answer to add some explanation, including the assumptions you've made.