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
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
n and sums all the elements with
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
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
1to 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 <= nThis 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_sumis 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.joinmethod. 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
- (cont) Similarly,
sortedcan accept a generator, but it has to construct a list, which it then calls the
list.sortmethod on, so there's no advantage in using
sortedon a generator, rather than
.sorton 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...
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