# 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 forth powers ` `// of first n natural numbers ` `#include ` `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

 ` `

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 ` `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

 ` `

Output

```2275
```

Time Complexity : O(1)

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