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 sumint calculateSum(int n){ // Return total sum return pow(n * (n + 1) / 2, 2);}// Driver codeint 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 sumdef calculateSum(n): #return total sum return (n * (n + 1) / 2)**2 #Driver codeif __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^3int 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: 100print(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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