## Sum of the integers from 1 to n

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find the sum of integers from 1 to 100
n(n+1)/2
sum of numbers from 1 to 99
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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
```

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

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.

• `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
• 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.