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Sum of first n odd numbers in O(1) Complexity

  • Difficulty Level : Easy
  • Last Updated : 24 May, 2021

Given the sequence of odd numbers 
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, …. 
Find the sum of first n odd numbers
Examples: 
 

Input : n = 2
Output : 4
Sum of first two odd numbers is 1 + 3 = 4.

Input : 5
Output : 25
Sum of first 5 odd numbers is 1 + 3 + 5 +
7 + 9 = 25

 

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A simple solution is to iterate through all odd numbers. 
 

C++




// A naive CPP program to find sum of
// first n odd numbers
#include <iostream>
using namespace std;
 
// Returns the sum of first
// n odd numbers
int oddSum(int n)
{
    int sum = 0, curr = 1;
    for (int i = 0; i < n; i++) {
        sum += curr;
        curr += 2;
    }
    return sum;
}
 
// Driver function
int main()
{
    int n = 20;
    cout << " Sum of first " << n
         << " Odd Numbers is: " << oddSum(n);
    return 0;
}

Java




// Java program to find sum of
// first n odd numbers
import java.util.*;
 
class Odd
{  
    // Returns the sum of first
    // n odd numbers
    public static int oddSum(int n)
    {
        int sum = 0, curr = 1;
        for (int i = 0; i < n; i++) {
            sum += curr;
            curr += 2;
        }
        return sum;
    }
     
    // driver function
    public static void main(String[] args)
    {
        int n = 20;
        System.out.println(" Sum of first "+ n
        +" Odd Numbers is: "+oddSum(n));
    }
}
 
// This code is contributed by rishabh_jain

Python3




# Python3 program to find sum
# of first n odd numbers
 
def oddSum(n) :
    sum = 0
    curr = 1
    i = 0
    while i < n:
        sum = sum + curr
        curr = curr + 2
        i = i + 1
    return sum
 
# Driver Code
n = 20
print (" Sum of first" , n, "Odd Numbers is: ",
                                oddSum(n) )
 
# This code is contributed by rishabh_jain

C#




// C# program to find sum of
// first n odd numbers
using System;
 
class GFG {
     
    // Returns the sum of first
    // n odd numbers
    public static int oddSum(int n)
    {
        int sum = 0, curr = 1;
        for (int i = 0; i < n; i++) {
            sum += curr;
            curr += 2;
        }
         
        return sum;
    }
 
    // driver function
    public static void Main()
    {
        int n = 20;
        Console.WriteLine(" Sum of first " + n
            + " Odd Numbers is: " + oddSum(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// A naive PHP program to find sum of
// first n odd numbers
 
// Returns the sum of first
// n odd numbers
function oddSum($n)
{
    $sum = 0; $curr = 1;
    for ($i = 0; $i < $n; $i++)
    {
        $sum += $curr;
        $curr += 2;
    }
    return $sum;
}
 
// Driver Code
$n = 20;
echo " Sum of first ", $n
     , " Odd Numbers is: ", oddSum($n);
      
// This code is contributed by vt_m.
?>

Javascript




<script>
 
// A naive Javascript program to find sum of
// first n odd numbers
 
// Returns the sum of first
// n odd numbers
function oddSum(n)
{
    let sum = 0; curr = 1;
    for (let i = 0; i < n; i++)
    {
        sum += curr;
        curr += 2;
    }
    return sum;
}
 
// Driver Code
let n = 20;
document.write(" Sum of first " + n
     + " Odd Numbers is: " + oddSum(n));
      
// This code is contributed by gfgking.
</script>

Output: 



Sum of first 20 odd numbers is 400

Time Complexity: O(n) 
Auxiliary Space : O(1)
 
An efficient solution is to use direct formula. To find the sum of first n odd numbers we can apply odd number theorem, it states that the sum of first n odd numbers is equal to the square of n.
 

∑(2i – 1) = n2 where i varies from 1 to n

let n = 10, therefore sum of first 10 odd numbers is
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
if we apply odd number theorem:
sum of first 10 odd numbers = n * n = 10 * 10 = 100.
Below is the implementation of the above approach: 
 

C++




// Efficient program to find sum of
// first n odd numbers
#include <iostream>
using namespace std;
 
// Returns the sum of first
// n odd numbers
int oddSum(int n)
{
    return (n * n);
}
 
// Driver function
int main()
{
    int n = 20;
    cout << " Sum of first " << n
         << " Odd Numbers is: " << oddSum(n);
    return 0;
}

Java




// Java program to find sum of
// first n odd numbers
import java.util.*;
 
class Odd
{  
    // Returns the sum of first
    // n odd numbers
    public static int oddSum(int n)
    {
        return (n * n);
    }
     
    // driver function
    public static void main(String[] args)
    {
        int n = 20;
        System.out.println(" Sum of first "+ n
        +" Odd Numbers is: "+oddSum(n));
    }
}
 
// This code is contributed by rishabh_jain

Python3




# Python3 program to find sum
# of first n odd numbers
 
def oddSum(n) :
    return (n * n);
 
# Driver Code
n = 20
print (" Sum of first" , n, "Odd Numbers is: ",
                               oddSum(n) )
 
# This code is contributed by rishabh_jain

C#




// C# program to find sum of
// first n odd numbers
using System;
 
class GFG {
     
    // Returns the sum of first
    // n odd numbers
    public static int oddSum(int n)
    {
        return (n * n);
    }
 
    // driver function
    public static void Main()
    {
        int n = 20;
        Console.WriteLine(" Sum of first " + n
            + " Odd Numbers is: " + oddSum(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// Efficient program to find sum of
// first n odd numbers
 
// Returns the sum of first
// n odd numbers
function oddSum($n)
{
    return ($n * $n);
}
 
// Driver Code
$n = 20;
echo " Sum of first " , $n,
     " Odd Numbers is: ", oddSum($n);
 
// This code is contributed by vt_m.
?>

Javascript




<script>
    // Javascript program to find sum of first n odd numbers
     
    // Returns the sum of first
    // n odd numbers
    function oddSum(n)
    {
        return (n * n);
    }
     
    let n = 20;
    document.write(" Sum of first " + n
                      + " Odd Numbers is: " + oddSum(n));
     
    // This code is contributed by divyesh072019.
</script>

Output: 
 

Sum of first 20 odd numbers is 400

Time Complexity: O(1) 
Auxiliary Space : O(1)
How does it work? 
We can prove it using mathematical induction. We know it is true for n = 1 and n = 2 as sums are 1 and 4 (1 + 3) respectively.
 

Let it be true for n = k-1.

Sum of first k odd numbers = 
  Sum of first k-1 odd numbers + k'th odd number
= (k-1)*(k-1) + (2k - 1)
= k*k

 




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