Sum of all natural numbers in range L to R

Given a range L and R, the task is to find the sum of all natural numbers in range L to R.

Examples:

Input: L = 2, R = 5
Output: 14
2 + 3 + 4 + 5 = 14

Input: L = 10, R = 20
Output: 165

A naive approach is to traverse from L to R and add all the elements one by one to get the sum.



An efficient approach is to use the formula for the sum of first N natural numbers. The idea of the inclusion-exclusion principle helps to solve the above problem. Find the sum of natural numbers till R and L-1 and then subtract sum(R)-sum(l-1).

Below is the implementation of the above approach:

C++

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// C++ program to print the sum
// of all numbers in range L and R
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the sum of
// all natural numbers
int sumNatural(int n)
{
    int sum = (n * (n + 1)) / 2;
    return sum;
}
  
// Function to return the sum
// of all numbers in range L and R
int suminRange(int l, int r)
{
    return sumNatural(r) - sumNatural(l - 1);
}
  
// Driver Code
int main()
{
    int l = 2, r = 5;
    cout << "Sum of Natural numbers from L to R is "
         << suminRange(l, r);
  
    return 0;
}

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Java

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// Java program to print the sum
// of all numbers in range L and R
  
class GFG{
// Function to return the sum of
// all natural numbers
static int sumNatural(int n)
{
    int sum = (n * (n + 1)) / 2;
    return sum;
}
  
// Function to return the sum
// of all numbers in range L and R
static int suminRange(int l, int r)
{
    return sumNatural(r) - sumNatural(l - 1);
}
  
// Driver Code
public static void main(String[] args)
{
    int l = 2, r = 5;
    System.out.println("Sum of Natural numbers from L to R is "+suminRange(l, r));
  
}
}
// This code is contributed by mits

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Python3

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# Python3 program to print the sum  of 
# all numbers in range L and R
  
# Function to return the sum of all natural numbers
def sumNatural(n):
  
    sum = (n*(n+1))//2
  
    return sum
  
# Function to return the sum
# of all numbers in range L and R
def suminRange(l, r):
    return sumNatural(r) - sumNatural(l-1)
  
#  Driver Code
l =2; r= 5
print("Sum of Natural numbers from L to R is ",suminRange(l, r))
  
# This code is contributed by Shrikant13

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C#

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// C# program to print the sum 
// of all numbers in range L and R 
using System;
  
class GFG
{
// Function to return the sum 
// of all natural numbers 
static int sumNatural(int n) 
    int sum = (n * (n + 1)) / 2; 
    return sum; 
  
// Function to return the sum 
// of all numbers in range L and R 
static int suminRange(int l, int r) 
    return sumNatural(r) - 
           sumNatural(l - 1); 
  
// Driver Code 
static public void Main ()
{
    int l = 2, r = 5; 
    Console.WriteLine("Sum of Natural numbers "
                              "from L to R is "
                               suminRange(l, r)); 
  
// This code is contributed by akt_mit

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PHP

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<?php
// PHP program to print the sum
// of all numbers in range L and R
  
// Function to return the sum of
// all natural numbers
function sumNatural($n)
{
    $sum = ($n * ($n + 1)) / 2;
    return $sum;
}
  
// Function to return the sum
// of all numbers in range L and R
function suminRange($l, $r)
{
    return sumNatural($r) - 
           sumNatural($l - 1);
}
  
// Driver Code
$l = 2;
$r = 5;
echo "Sum of Natural numbers "
              "from L to R is ",
             suminRange($l, $r);
  
// This code is contributed by ajit
?>

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Output:

Sum of Natural numbers from L to R is 14


My Personal Notes arrow_drop_up

Just another competitive programmer

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