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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

PHP

<?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
?>

Javascript

<script>
// JavaScript 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
let l = 2;
let r = 5;
document.write("Sum of Natural numbers from L to R is "+
             suminRange(l, r));

// This code is contributed by sravan kumar gottumukkalan

</script>

Output:

Sum of Natural numbers from L to R is 14

Time Complexity: O(1)

Auxiliary Space: O(1), since no extra space has been taken.