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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