C++ Program to Calculate Sum of Natural Numbers

Using For-Loop

From 1 to n, a for loop is executed. Iterator'  i' represents the natural numbers from 1 to n. So, this value of i will be added to the sum during each iteration. Therefore, the sum of first 'n' Natural numbers is calculated using for loop. The following code sample demonstrates this.

#include<iostream>
using namespace std;
int main() 
{
   int n=10, sum=0, i;
   for(i=1;i<=n;i++)
   {
     sum=sum+i;
   }
   cout<<"Sum of first "<<n<<" natural numbers is "<<sum;
   return 0;
}

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Output

Sum of first 10 natural numbers is 55

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Time Complexity: O(n) and we run loop for n number and find the total sum.

Space Complexity: O(1)

Using Formula

The formula used to find the sum of first n natural numbers, is given as:

sum = n(n+1)/2

The following is a program that uses the above formula to calculate the sum of n natural numbers.

#include<iostream>
using namespace std;
int main() {
   int n=10, sum=0;
   sum = n*(n+1)/2;
   cout<<"Sum of first "<<n<<" natural numbers is "<<sum;
   return 0;
}

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Output

Sum of first 5 natural numbers is 15

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Time Complexity: O(1)

Space Complexity: O(1)

Try and compile with online c++ compiler.

Must Read Dynamic Binding in C++

Also Read - Strong number in c

FAQs

1. What is the complexity of summing a series of n natural numbers?
   In this case, algorithm's complexity is calculated as the number of times addition operation is performed that is O(n).
2. Why do we use #include iostream in C++ programming?
   Iostream is a file that defines all of the commands such as cout, endl, etc. We'll need to add that file if we want to use them. #include iostream> simply implies copying and pasting the code from that file into your code.

Key Takeaways

This article extensively discussed the program to calculate the sum of n natural numbers using C++. The article explains the different approaches used to find sum of natural numbers in C++. The use of formula as well as loops are explained. There are many more approaches, but these are the most efficient in terms of time and space.

Check out this problem - Two Sum Problem

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