Program to find Sum of the series 1*3 + 3*5 + ....

Given a series:

S_n = 1*3 + 3*5 + 5*7 + …

It is required to find the sum of first n terms of this series represented by S_n, where n is given taken input.

Examples:

Input : n = 2 
Output : S<sub>n</sub> = 18
Explanation:
The sum of first 2 terms of Series is
1*3 + 3*5
= 3 + 15 
= 28

Input : n = 4 
Output : S<sub>n</sub> = 116
Explanation:
The sum of first 4 terms of Series is
1*3 + 3*5 + 5*7 + 7*9
= 3 + 15 + 35 + 63
= 116

Let, the n-th term be denoted by t_n.
This problem can easily be solved by observing that the nth term can be founded by following method:

t_n = (n-th term of (1, 3, 5, … ) )*(nth term of (3, 5, 7, ….))

Now, n-th term of series 1, 3, 5 is given by 2*n-1
and, the n-th term of series 3, 5, 7 is given by 2*n+1

Putting these two values in t_n:

t_n = (2*n-1)*(2*n+1) = 4*n*n-1

Now, the sum of first n terms will be given by :

S_n = ∑(4*n*n – 1)
=∑4*{n*n}-∑(1)

Now, it is known that the sum of first n terms of series n*n (1, 4, 9, …) is given by: n*(n+1)*(2*n+1)/6
And sum of n number of 1’s is n itself.

Now, putting values in S_n:

S_n = 4*n*(n+1)*(2*n+1)/6 – n
= n*(4*n*n + 6*n – 1)/3

Now, S_n value can be easily found by putting the desired value of n.

Below is the implementation of the above approach:

C++

// C++ program to find sum of first n terms 
#include <bits/stdc++.h> 
using namespace std; 
  
int calculateSum(int n) 
{ 
    // Sn = n*(4*n*n + 6*n - 1)/3 
    return (n * (4 * n * n + 6 * n - 1) / 3); 
} 
  
int main() 
{ 
    // number of terms to be included in the sum 
    int n = 4; 
  
    // find the Sn 
    cout << "Sum = " << calculateSum(n); 
  
    return 0; 
} 

Java

// Java program to find sum  
// of first n terms  
class GFG  
{ 
    static int calculateSum(int n)  
    {  
        // Sn = n*(4*n*n + 6*n - 1)/3  
        return (n * (4 * n * n +  
                     6 * n - 1) / 3);  
    }  
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        // number of terms to be  
        // included in the sum  
        int n = 4;  
      
        // find the Sn  
        System.out.println("Sum = " +  
                            calculateSum(n)); 
    } 
} 
  
// This code is contributed by Bilal 

Python

# Python program to find sum  
# of first n terms  
def calculateSum(n):  
      
    # Sn = n*(4*n*n + 6*n - 1)/3  
    return (n * (4 * n * n + 
                 6 * n - 1) / 3); 
  
# Driver Code 
  
# number of terms to be  
# included in the sum  
n = 4
  
# find the Sn  
print("Sum =",calculateSum(n)) 
  
# This code is contributed by Bilal 

C#

// C# program to find sum  
// of first n terms  
using System; 
  
class GFG 
{ 
  
static int calculateSum(int n)  
{  
    // Sn = n*(4*n*n + 6*n - 1)/3  
    return (n * (4 * n * n +  
                 6 * n - 1) / 3);  
}  
  
// Driver code 
static public void Main () 
{ 
    // number of terms to be  
    // included in the sum  
    int n = 4;  
  
    // find the Sn  
    Console.WriteLine("Sum = " +  
                       calculateSum(n)); 
} 
} 
  
// This code is contributed 
// by mahadev 

PHP

<?php  
// PHP program to find sum  
// of first n terms 
  
function calculateSum($n) 
{ 
    // Sn = n*(4*n*n + 6*n - 1)/3 
    return ($n * (4 * $n * $n +  
                  6 * $n - 1) / 3); 
} 
  
// number of terms to be 
// included in the sum 
$n = 4; 
  
// find the Sn 
echo "Sum = " . calculateSum($n); 
  
// This code is contributed 
// by ChitraNayal 
?> 

Output:

Sum = 116

My Personal Notes

Program to find Sum of the series 13 + 35 + ….

C++

Java

Python

C#

PHP

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