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.