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 opf 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; } |
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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.. |
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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 |
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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) |
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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 ?> |
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Output:
100
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