Find the sum of n terms of the series 1,8,27,64 ….

Given a series, the task is to find the sum of the below series up to n terms:

1, 8, 27, 64, …

Examples:

Input: N = 2
Output: 9
9 = (2*(2+1)/2)^2
Input: N = 4
Output: 100
100 = (4*(4+1)/2)^2

Approach: We can solve this problem using the following formula: 

    Sn = 1 + 8 + 27 + 64 + .........up to n terms
    Sn = (n*(n+1)/2)^2

Below is the implementation of above approach: 

C++

// C++ program to find the sum of n terms
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the sum
int calculateSum(int n)
{

    // Return total sum
    return pow(n * (n + 1) / 2, 2);
}

// Driver code
int main()
{

    int n = 4;
    cout << calculateSum(n);

    return 0;
}

Java

// Java program to find the sum of n terms
import java.io.*;

class GFG {


// Function to calculate the sum
static int calculateSum(int n)
{

    // Return total sum
    return (int)Math.pow(n * (n + 1) / 2, 2);
}

// Driver code


    public static void main (String[] args) {
        int n = 4;
    System.out.println( calculateSum(n));

    }
}
// This code is contributed by inder_verma..

Python3

# Python3 program to find the
# sum of n terms

#Function to calculate the sum
def calculateSum(n):

    #return total sum
    return (n * (n + 1) / 2)**2

#Driver code
if __name__=='__main__':
    n = 4
    print(calculateSum(n))

#this code is contributed by Shashank_Sharma

C#

// C# program to find the sum of n terms
using System;

class GFG
{

// Function ot calculate the sum
static int calculateSum(int n)
{

    // Return total sum
    return (int)Math.Pow(n * (n + 1) / 2, 2);
}

// Driver code
public static void Main ()
{
    int n = 4;
    Console.WriteLine(calculateSum(n));
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

<script>

// javascript program to find the sum of n terms

// Function to calculate the sum
function calculateSum(n)
{

    // Return total sum
    return parseInt(Math.pow(n * (n + 1) / 2, 2));
}

// Driver code
 var n = 4;
document.write( calculateSum(n));


// This code contributed by shikhasingrajput

</script>

PHP

<?php
// PHP program to find
// the sum of n terms

// Function to calculate the sum
function calculateSum($n)
{
    // Return total sum
    return pow($n * ($n + 1) / 2 , 2);
}

// Driver code
$n = 4;
echo calculateSum($n);

// This code is contributed
// by ANKITRAI1
?>

Output

100

Time complexity: O(logn) because using inbuilt function pow

Auxiliary Space: O(1)

 Using Loop:

Approach:

• Define a function sum_of_series_1 that takes an integer n as input.
• Initialize a variable sum to 0.
• Use a for loop that iterates over the range of n values.
• Inside the loop, add the cube of (i+1) to the variable sum.
• Return the value of sum after the loop completes.

C++

#include <iostream>

// Function to calculate the sum of the series 1^3 + 2^3 +
// 3^3 + ... + n^3
int sumOfSeries(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += (i + 1) * (i + 1)
               * (i + 1); // Calculate the cube of each
                          // number and add to the sum
    }
    return sum;
}

int main()
{
    std::cout << sumOfSeries(4) << std::endl; // Output: 100
    std::cout << sumOfSeries(2) << std::endl; // Output: 9
    return 0;
}

Java

public class SumOfSeries {

    // Function to calculate the sum of the series 1^3 + 2^3
    // + 3^3 + ... + n^3
    static int sumOfSeries(int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += Math.pow(
                (i + 1), 3); // Calculate the cube of each
                             // number and add to the sum
        }
        return sum;
    }

    public static void main(String[] args)
    {
        System.out.println(sumOfSeries(4)); // Output: 100
        System.out.println(sumOfSeries(2)); // Output: 9
    }
}

Python3

def sum_of_series_1(n):
    sum = 0
    for i in range(n):
        sum += (i+1)**3
    return sum

# Example usage:
print(sum_of_series_1(4))  # Output: 100
print(sum_of_series_1(2))  # Output: 9

C#

using System;

class Program {
    // Function to calculate the sum of the series 1^3 + 2^3
    // + 3^3 + ... + n^3
    static int SumOfSeries(int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += (int)Math.Pow(
                i + 1, 3); // Calculate the cube of each
                           // number and add to the sum
        }
        return sum;
    }

    static void Main()
    {
        Console.WriteLine(SumOfSeries(4)); // Output: 100
        Console.WriteLine(SumOfSeries(2)); // Output: 9
    }
}

Javascript

// Function to calculate the sum of the series 1^3 + 2^3 + 3^3 + ... + n^3
function sumOfSeries(n) {
    let sum = 0;
    for (let i = 0; i < n; i++) {
        sum += Math.pow(i + 1, 3); // Calculate the cube of each number and add to the sum
    }
    return sum;
}

// Test the function
console.log(sumOfSeries(4)); // Output: 100
console.log(sumOfSeries(2)); // Output: 9

Output

100
9

The time complexity of this approach is O(n) because we use a loop that iterates over n values.
The space complexity is O(1) because we use only one variable sum.