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

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

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

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 
?> 

Output:

Sum = 55

Efficient Approach:
Let a_n be the n-th term of the given series.

a_n = (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 
?> 

Output:

Sum = 55

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

Python3

C#

php

C++

Java

Python3

C#

PHP

Recommended Posts: