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

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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

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;
}

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

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.

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