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Sum of fourth powers of the first n natural numbers

  • Last Updated : 31 Mar, 2021

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.
?>

Javascript




<script>
 
// javascript Program to find the sum of forth powers
// of first n natural numbers
 
// Return the sum of forth 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

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.
?>

Javascript




<script>
 
// javascript Program to find the
// sum of forth powers of
// first n natural numbers
 
    
// Return the sum of
// forth 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

Time Complexity : O(1)
 

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