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++ 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 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 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# 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

<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>

<?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
?>

Sum = 116

Time Complexity: O(1)

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

METHOD 2:Using list comprehension .

APPROACH:

This program calculates the sum of the series 1*3 + 3*5 + ... using a list comprehension to generate the terms of the series and then finding their sum. The input value n determines the number of terms in the series to be generated and added. The sum of the series is then printed as the output.

ALGORITHM:

1.Take input value for n.
2.Generate the series using a list comprehension and store it in the series list.
3.Calculate the sum of the series list using the sum() function and store it in the variable sum.
4.Print the value of sum as the output.

#include <iostream>
#include <vector>

int main() {
    int n = 4;
    std::vector<int> series;

    // Generate the series using list comprehension
    for (int i = 0; i < n; i++) {
        int term = (2 * i + 1) * (2 * i + 3);
        series.push_back(term);
    }

    // Calculate the sum of the series
    int sum = 0;
    for (int i = 0; i < series.size(); i++) {
        sum += series[i];
    }

    // Print the sum
    std::cout << "Sum of the series: " << sum << std::endl;

    return 0;
}
// This code is contributed by uomkar369

import java.util.ArrayList;
import java.util.List;

public class Main {

    /**
     * This program generates the series of 2 * i + 1 * (2 * i + 3) for i in range(0, n).
     *
     * @param n The number of terms in the series.
     */
    public static void main(String[] args) {
        int n = 4; // The number of terms in the series.

        // Generate the series.
        List<Integer> series = new ArrayList<>(); // A list to store the terms of the series.
        for (int i = 0; i < n; i++) { // Iterate over the number of terms.
            int term = (2 * i + 1) * (2 * i + 3); // Calculate the term of the series.
            series.add(term); // Add the term to the list.
        }

        // Calculate the sum of the series.
        int sum = 0; // The sum of the series.
        for (int i = 0; i < series.size(); i++) { // Iterate over the list of terms.
            sum += series.get(i); // Add the current term to the sum.
        }

        // Print the sum.
        System.out.println("Sum of the series: " + sum); // Print the sum of the series.
    }
}

# Using a list comprehension
n = 4
series = [(2*i+1)*(2*i+3) for i in range(n)]
sum = sum(series)
print("Sum of the series:", sum)

using System;
using System.Collections.Generic;

class GFG
{
    static void Main()
    {
        int n = 4;
        List<int> series = new List<int>();

        // Generate the series using list comprehension
        for (int i = 0; i < n; i++)
        {
            int term = (2 * i + 1) * (2 * i + 3);
            series.Add(term);
        }

        // Calculate the sum of the series
        int sum = 0;
        foreach (int item in series)
        {
            sum += item;
        }

        // Print the sum
        Console.WriteLine("Sum of the series: " + sum);
    }
}
// This code is contributed by uomkar369

let n = 4;
let series = new Array();

// Generate the series using list comprehension
for (let i = 0; i < n; i++) {
    let term = (2 * i + 1) * (2 * i + 3);
    series.push(term);
}

// Calculate the sum of the series
let sum = 0;
for (let i = 0; i < series.length; i++) {
    sum += series[i];
}

// Prlet the sum
document.write("Sum of the series: " + sum);

Sum of the series: 116

Time Complexity:
The time complexity of this program is O(n), where n is the input value. This is because the program generates n terms of the series and then calculates their sum using the sum() function, which has a time complexity of O(n).

Space Complexity:
The space complexity of this program is also O(n), where n is the input value. This is because the program generates n terms of the series and stores them in the series list, which has a space complexity of O(n). The variable sum also requires constant space, so it does not affect the space complexity.