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 
= 18

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
?>

Javascript

<script>

// Javascript 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);
    }

    // Driver Code

        // number of terms to be
        // included in the sum
        let n = 4;

        // find the Sn
        document.write("Sum = " + calculateSum(n));

// This code contributed by Princi Singh

</script>

Output:

Sum = 116

Time Complexity: O(1)

Auxiliary Space: O(1), since no extra space has been taken.