Sum of fourth powers of the first n natural numbers
Write a program to find the sum of fourth powers of the first n natural numbers 14 + 24 + 34 + 44 + …….+ n4 till n-th term.
Examples :
Input : 4 Output : 354 14 + 24 + 34 + 44 = 354 Input : 6 Output : 2275 14 + 24 + 34 + 44+ 54+ 64 = 2275
Naive Approach :- Simple finding the fourth powers of the first n natural numbers is iterate a loop from 1 to n time. like suppose n=4.
(1*1*1*1)+(2*2*2*2)+(3*3*3*3)+(4*4*4*4) = 354
C++
// CPP Program to find the sum of fourth powers // of first n natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of fourth power of first n // natural numbers long long int fourthPowerSum( int n) { long long int sum = 0; for ( int i = 1; i <= n; i++) sum = sum + (i * i * i * i); return sum; } // Driven Program int main() { int n = 6; cout << fourthPowerSum(n) << endl; return 0; } |
Java
// Java Program to find the // sum of fourth powers of // first n natural numbers import java.io.*; import java.util.*; class GFG { // Return the sum of fourth // power of first n natural // numbers static long fourthPowerSum( int n) { long sum = 0 ; for ( int i = 1 ; i <= n; i++) sum = sum + (i * i * i * i); return sum; } public static void main (String[] args) { int n = 6 ; System.out.println(fourthPowerSum(n)); } } // This code is contributed by Gitanjali. |
Python3
# Python3 Program to find the # sum of fourth powers of first # n natural numbers import math # Return the sum of fourth power of # first n natural numbers def fourthPowerSum( n): sum = 0 for i in range ( 1 , n + 1 ) : sum = sum + (i * i * i * i) return sum # Driver method n = 6 print (fourthPowerSum(n)) # This code is contributed by Gitanjali. |
C#
// C# program to find the // sum of fourth powers of // first n natural numbers using System; class GFG { // Return the sum of fourth power // of first n natural numbers static long fourthPowerSum( int n) { long sum = 0; for ( int i = 1; i <= n; i++) sum = sum + (i * i * i * i); return sum; } public static void Main () { int n = 6; Console.WriteLine(fourthPowerSum(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP Program to find th // sum of fourth powers // of first n natural numbers // Return the sum of fourth // power of first n // natural numbers function fourthPowerSum( $n ) { $sum = 0; for ( $i = 1; $i <= $n ; $i ++) $sum = $sum + ( $i * $i * $i * $i ); return $sum ; } // Driver Code $n = 6; echo (fourthPowerSum( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // javascript Program to find the sum of fourth powers // of first n natural numbers // Return the sum of fourth power of first n // natural numbers function fourthPowerSum( n) { let sum = 0; for (let i = 1; i <= n; i++) sum = sum + (i * i * i * i); return sum; } // Driven Program let n = 6; document.write(fourthPowerSum(n)); // This code contributed by aashish1995 </script> |
Output:
2275
Complexity Analysis:
Time Complexity : O(n) ,as there is single loop used inside fourthpowersum() function.
Space Complexity: O(1) , as there is no extra space used.
Efficient Approach :- An efficient solution is to use direct mathematical formula which is 1/30n(n+1)(2n+1)(3n2+3n+1) or it is also write (1/5)n5 + (1/2)n4 + (1/3)n3 – (1/30)n. This solution take O(1) time.
C++
// CPP Program to find the sum of fourth power of first // n natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of fourth power of first n natural // numbers long long int fourthPowerSum( int n) { return ((6 * n * n * n * n * n) + (15 * n * n * n * n) + (10 * n * n * n) - n) / 30; } // Driven Program int main() { int n = 6; cout << fourthPowerSum(n) << endl; return 0; } |
Java
// Java Program to find the // sum of fourth powers of // first n natural numbers import java.io.*; import java.util.*; class GFG { // Return the sum of // fourth power of first // n natural numbers static long fourthPowerSum( int n) { return (( 6 * n * n * n * n * n) + ( 15 * n * n * n * n) + ( 10 * n * n * n) - n) / 30 ; } public static void main (String[] args) { int n = 6 ; System.out.println(fourthPowerSum(n)); } } // This code is contributed by Gitanjali. |
Python3
# Python3 Program to # find the sum of # fourth powers of # first n natural numbers import math # Return the sum of # fourth power of # first n natural # numbers def fourthPowerSum(n): return (( 6 * n * n * n * n * n) + ( 15 * n * n * n * n) + ( 10 * n * n * n) - n) / 30 # Driver method n = 6 print (fourthPowerSum(n)) # This code is contributed by Gitanjali. |
C#
// C# Program to find the // sum of fourth powers of // first n natural numbers using System; class GFG { // Return the sum of // fourth power of first // n natural numbers static long fourthPowerSum( int n) { return ((6 * n * n * n * n * n) + (15 * n * n * n * n) + (10 * n * n * n) - n) / 30; } public static void Main () { int n = 6; Console.Write(fourthPowerSum(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP Program to find the sum // of fourth power of first // n natural numbers // Return the sum of fourth // power of first n natural // numbers function fourthPowerSum( $n ) { return ((6 * $n * $n * $n * $n * $n ) + (15 * $n * $n * $n * $n ) + (10 * $n * $n * $n ) - $n ) / 30; } // Driver Code $n = 6; echo (fourthPowerSum( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // javascript Program to find the // sum of fourth powers of // first n natural numbers // Return the sum of // fourth power of first // n natural numbers function fourthPowerSum(n) { return ((6 * n * n * n * n * n) + (15 * n * n * n * n) + (10 * n * n * n) - n) / 30; } var n = 6; document.write(fourthPowerSum(n)); // This code is contributed by 29AjayKumar </script> |
Output:
2275
Complexity Analysis:
Time Complexity : O(1)
Space Complexity: O(1)
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