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Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + …… + (1+3+5+7+…+(2n-1))

  • Difficulty Level : Easy
  • Last Updated : 20 Apr, 2021

Given a positive integer n. The problem is to find the sum of the given series 1 + (1+2) + (1+2+3) + (1+2+3+4) + …… + (1+2+3+4+…+n), where i-th term in the series is the sum of first i odd natural numbers.
Examples: 
 

Input : n = 2
Output : 5
(1) + (1+3) = 5

Input : n = 5
Output : 55
(1) + (1+3) + (1+3+5) + (1+3+5+7) + (1+3+5+7+9) = 55

 

Naive Approach: Using two loops get the sum of each i-th term and then add those sum to the final sum.
 

C++




// C++ implementation to find the
// sum of the given series
#include <bits/stdc++.h>
 
using namespace std;
 
// functionn to find the
// sum of the given series
int sumOfTheSeries(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; i++) {
 
        // first term of each i-th term
        int k = 1;
        for (int j = 1; j <= i; j++) {
            sum += k;
 
            // next term
            k += 2;
        }
    }
 
    // required sum
    return sum;
}
 
// Driver program
int main()
{
    int n = 5;
    cout << "Sum = "
         << sumOfTheSeries(n);
    return 0;
}
Java



// Java implementation to find
// the sum of the given series
import java.util.*;
 
class GFG {
     
    // functionn to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
        {
      
            // first term of each
            // i-th term
            int k = 1;
            for (int j = 1; j <= i; j++)
            {
                sum += k;
      
                // next term
                k += 2;
            }
        }
      
        // required sum
        return sum;
    }
 
    /* Driver program */
    public static void main(String[] args)
    {
         int n = 5;
         System.out.println("Sum = " +
                        sumOfTheSeries(n));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.
Python3



# Python3 implementation to find
# the sum of the given series
 
# functionn to find the sum
# of the given series
def sumOfTheSeries( n ):
    sum = 0
    for i in range(1, n + 1):
         
        # first term of each i-th term
        k = 1
        for j in range(1,i+1):
            sum += k
             
            # next term
            k += 2
             
    # required sum
    return sum
     
# Driver program
n = 5
print("Sum =", sumOfTheSeries(n))
 
# This code is contributed by "Sharad_Bhardwaj".
C#



// C# implementation to find
// the sum of the given series
using System;
 
class GFG {
 
    // functionn to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
 
            // first term of each
            // i-th term
            int k = 1;
            for (int j = 1; j <= i; j++) {
                sum += k;
 
                // next term
                k += 2;
            }
        }
 
        // required sum
        return sum;
    }
 
    /* Driver program */
    public static void Main()
    {
        int n = 5;
        Console.Write("Sum = " +
                     sumOfTheSeries(n));
    }
}
 
// This code is contributed by vt_m.
php



<?php
 
// php implementation to find the
// sum of the given series
 
// functionn to find the
// sum of the given series
function sumOfTheSeries($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++) {
  
        // first term of each i-th term
        $k = 1;
        for ($j = 1; $j <= $i; $j++) {
            $sum += $k;
  
            // next term
            $k += 2;
        }
    }
  
    // required sum
    return $sum;
}
  
// Driver program
    $n = 5;
    echo "Sum = "
         . sumOfTheSeries($n);
 
// This code is contributed by Sam007
?>
Javascript



<script>
 
// Javascript implementation to find the
// sum of the given series
 
// functionn to find the
// sum of the given series
function sumOfTheSeries(n)
{
    let sum = 0;
    for (let i = 1; i <= n; i++) {
  
        // first term of each i-th term
        let k = 1;
        for (let j = 1; j <= i; j++) {
            sum += k;
  
            // next term
            k += 2;
        }
    }
  
    // required sum
    return sum;
}
  
// Driver program
    let n = 5;
    document.write("Sum = " + sumOfTheSeries(n));
 
// This code is contributed by gfgking
 
</script>
Output: 
 
Sum = 55
Efficient Approach: 
Let an be the n-th term of the given series. 
 


an = (1 + 3 + 5 + 7 + (2n-1))
   = sum of first n odd numbers
   = n2
Refer this post for the proof of above formula.
Now, 
 
Refer this post for the proof of above formula.
 
C++



// C++ implementation to find the sum
// of the given series
#include <bits/stdc++.h>
using namespace std;
 
// functionn to find the sum
// of the given series
int sumOfTheSeries(int n)
{
    // required sum
    return (n * (n + 1) / 2) *
            (2 * n + 1) / 3;
}
 
// Driver program to test above
int main()
{
    int n = 5;
    cout << "Sum = "
         << sumOfTheSeries(n);
    return 0;
}
Java



// Java implementation to find
// the sum of the given series
import java.io.*;
 
class GfG {
     
// function to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
    // required sum
    return (n * (n + 1) / 2) *
            (2 * n + 1) / 3;
}
     
 
// Driver program to test above
public static void main (String[] args)
{
    int n = 5;
     
    System.out.println("Sum = "+
                sumOfTheSeries(n));
 
}
 
}
 
// This code is contributed by Gitanjali.
Python3



# Python3 implementation to find
# the sum of the given series
 
# functionn to find the sum
# of the given series
def sumOfTheSeries( n ):
     
    # required sum
    return int((n * (n + 1) / 2) *
            (2 * n + 1) / 3)
             
# Driver program to test above
n = 5
print("Sum =", sumOfTheSeries(n))
 
# This code is contributed by "Sharad_Bhardwaj".
C#



// C# implementation to find
// the sum of the given series
using System;
 
class GfG {
 
    // function to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
        // required sum
        return (n * (n + 1) / 2) *
                      (2 * n + 1) / 3;
    }
 
    // Driver program to test above
    public static void Main()
    {
        int n = 5;
 
        Console.Write("Sum = " +
                   sumOfTheSeries(n));
    }
}
 
// This code is contributed by vt_m.
PHP



<?php
// PHP implementation to find the sum
// of the given series
 
// functionn to find the sum
// of the given series
function sumOfTheSeries($n)
{
     
    // required sum
    return ($n * ($n + 1) / 2) *
              (2 * $n + 1) / 3;
}
 
    // Driver Code
    $n = 5;
    echo "Sum = "
        . sumOfTheSeries($n);
         
// This code is contributed by Sam007
?>
Javascript



<script>
 
// JavaScript program to find
// the sum of the given series
 
// function to find the sum
    // of the given series
    function sumOfTheSeries(n)
    {
        // required sum
        return (n * (n + 1) / 2) *
                      (2 * n + 1) / 3;
    }
 
// Driver Code
 
        let n = 5;
   
        document.write("Sum = " +
                   sumOfTheSeries(n));
         
        // This code is contributed by avijitmondal1998.
</script>
Output: 
 
Sum = 55
 
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Article Contributed By :
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ayushjauhari14
@ayushjauhari14
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