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C++ "Hello, World!" ProgramPrint Number Entered by UserAdd Two NumbersFind Quotient and RemainderFind Size of int, float, double and char in Your SystemSwap Two NumbersFind ASCII Value of a CharacterMultiply two Numbers
Decisions and Loops
Check Whether Number is Even or OddCheck Whether a character is Vowel or ConsonantFind Largest Number Among Three NumbersFind All Roots of a Quadratic EquationCalculate Sum of Natural NumbersCheck Leap YearFind FactorialGenerate Multiplication TableDisplay Fibonacci SeriesFind GCDFind LCMReverse a NumberCalculate Power of a NumberCheck Whether a Number is Palindrome or NotCheck Whether a Number is Prime or NotDisplay Prime Numbers Between Two IntervalsCheck Armstrong NumberDisplay Armstrong Number Between Two IntervalsDisplay Factors of a NumberCreate Pyramid and PatternMake a Simple Calculator to Add, Subtract, Multiply or Divide Using switch...case
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Display Prime Numbers Between Two Intervals Using FunctionsCheck Prime Number By Creating a FunctionCheck Whether a Number can be Express as Sum of Two Prime NumbersFind Sum of Natural Numbers using RecursionCalculate Factorial of a Number Using RecursionFind G.C.D Using RecursionConvert Binary Number to Decimal and vice-versaConvert Octal Number to Decimal and vice-versaConvert Binary Number to Octal and vice-versaReverse a Sentence Using RecursionCalculate Power Using Recursion
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Increment ++ and Decrement -- Operator Overloading in C++ ProgrammingSubtract Complex Number Using Operator Overloading

C++ Program to Calculate Sum of Natural Numbers

To understand this example, you should have the knowledge of the following C++ programming topics:

Positive integers 1, 2, 3, 4… are known as natural numbers.

This program takes a positive integer from user (suppose user entered n ) then, this program displays the value of 1 + 2 + 3 + … + n.

Example: Sum of Natural Numbers using loop


              

#include <iostream>
using namespace std;

int main() {
    int n, sum = 0;

    cout << "Enter a positive integer: ";
    cin >> n;

    for (int i = 1; i <= n; ++i) {
        sum += i;
    }

    cout << "Sum = " << sum;
    return 0;
}

Output


              

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.