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

Given a series:

Sn = 1*3 + 3*5 + 5*7 + …

It is required to find the sum of first n terms of this series represented by Sn, 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 tn.
This problem can easily be solved by observing that the nth term can be founded by following method:

tn = (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 tn:

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

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

Sn = ∑(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 Sn:

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

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

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

Sum = 116


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Improved By : bilal-hungund, Mahadev99, Ita_c