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Sum of fourth powers of the first n natural numbers
  • Last Updated : 25 Apr, 2018

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 forth powers
// of first n natural numbers
#include <bits/stdc++.h>
using namespace std;
  
// Return the sum of forth 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 forth powers of 
// first n natural numbers
import java.io.*;
import java.util.*;
  
class GFG {
      
    // Return the sum of forth
    // 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 forth powers of first
# n natural numbers
import math 
  
# Return the sum of forth 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 forth powers of 
// first n natural numbers
using System;
  
class GFG {
      
    // Return the sum of forth 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.
?>


output

2275

Time Complexity : O(n)



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 forth power of first
// n natural numbers
#include <bits/stdc++.h>
using namespace std;
  
// Return the sum of forth 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 forth powers of
// first n natural numbers
import java.io.*;
import java.util.*;
  
class GFG {
      
    // Return the sum of 
    // forth 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 
# forth powers of 
# first n natural numbers
import math 
  
# Return the sum of 
# forth 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 forth powers of
// first n natural numbers
using System;
  
class GFG {
      
    // Return the sum of 
    // forth 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.
?>



Output

2275

Time Complexity : O(1)

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