Sum of fourth powers of the first n natural numbers
Last Updated :
19 Jan, 2023
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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int n = 6;
cout << fourthPowerSum(n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
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));
}
}
|
Python3
import math
def fourthPowerSum( n):
sum = 0
for i in range ( 1 , n + 1 ) :
sum = sum + (i * i * i * i)
return sum
n = 6
print (fourthPowerSum(n))
|
C#
using System;
class GFG {
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));
}
}
|
PHP
<?php
function fourthPowerSum( $n )
{
$sum = 0;
for ( $i = 1; $i <= $n ; $i ++)
$sum = $sum + ( $i * $i * $i * $i );
return $sum ;
}
$n = 6;
echo (fourthPowerSum( $n ));
?>
|
Javascript
<script>
function fourthPowerSum( n)
{
let sum = 0;
for (let i = 1; i <= n; i++)
sum = sum + (i * i * i * i);
return sum;
}
let n = 6;
document.write(fourthPowerSum(n));
</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++
#include <bits/stdc++.h>
using namespace std;
long long int fourthPowerSum( int n)
{
return ((6 * n * n * n * n * n) +
(15 * n * n * n * n) +
(10 * n * n * n) - n) / 30;
}
int main()
{
int n = 6;
cout << fourthPowerSum(n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
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));
}
}
|
Python3
import math
def fourthPowerSum(n):
return (( 6 * n * n * n * n * n) +
( 15 * n * n * n * n) +
( 10 * n * n * n) - n) / 30
n = 6
print (fourthPowerSum(n))
|
C#
using System;
class GFG {
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));
}
}
|
PHP
<?php
function fourthPowerSum( $n )
{
return ((6 * $n * $n * $n * $n * $n ) +
(15 * $n * $n * $n * $n ) +
(10 * $n * $n * $n ) - $n ) / 30;
}
$n = 6;
echo (fourthPowerSum( $n ));
?>
|
Javascript
<script>
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));
</script>
|
Output:
2275
Complexity Analysis:
Time Complexity : O(1)
Space Complexity: O(1)
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