# n-th term of series 1, 17, 98, 354……

Given a series 1, 17, 98, 354 …… Find the nth term of this series.

The series basically represents sum of 4th power of first n natural numbers. First term is sum of 14. Second term is sum of two numbers i.e (14 + 24 = 17), third term i.e.(14 + 24 + 34 = 98 ) and so on.

Examples :

```Input : 5
Output : 979

Input : 7
Output : 4676
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach :
A simple solution is to add the 4th powers of first n natural numbers. By using iteration we can easily find nth term of the series.

Below is the implementation of the above approach :

## C++

 `// CPP program to find n-th term of ` `// series ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the nth term of series ` `int` `sumOfSeries(``int` `n) ` `{ ` `    ``// Loop to add 4th powers ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``ans += i * i * i * i; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << sumOfSeries(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find n-th term of ` `// series ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the nth term of series ` `    ``static` `int` `sumOfSeries(``int` `n) ` `    ``{ ` `        ``// Loop to add 4th powers ` `        ``int` `ans = ``0``; ` `        ``for` `(``int` `i = ``1``; i <= n; i++) ` `            ``ans += i * i * i * i; ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(sumOfSeries(n)); ` `    ``} ` `} `

## Python3

 `# Python 3 program to find ` `# n-th term of ` `# series ` `  `  `      `  `# Function to find the ` `# nth term of series ` `def` `sumOfSeries(n) : ` `    ``# Loop to add 4th powers ` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``1``, n ``+` `1``) : ` `        ``ans ``=` `ans ``+` `i ``*` `i ``*` `i ``*` `i  ` `       `  `    ``return` `ans ` `  `  `  `  `# Driver code ` `n ``=` `4` `print``(sumOfSeries(n)) `

## C#

 `// C# program to find n-th term of ` `// series ` `using` `System; ` `class` `GFG { ` ` `  `    ``// Function to find the  ` `    ``// nth term of series ` `    ``static` `int` `sumOfSeries(``int` `n) ` `    ``{ ` `         `  `        ``// Loop to add 4th powers ` `        ``int` `ans = 0; ` `        ``for` `(``int` `i = 1; i <= n; i++) ` `            ``ans += i * i * i * i; ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4; ` `        ``Console.WriteLine(sumOfSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67 `

## PHP

 ` `

Output :

```354
```

Time Complexity : O(n).

Efficient approach :
The pattern in this series is nth term is equal to sum of (n-1)th term and n4.

Examples :

```n = 2
2nd term equals to sum of 1st term and 24 i.e 16
A2 = A1 + 16
= 1 + 16
= 17

Similarly,
A3 = A2 + 34
= 17 + 81
= 98 and so on..
```

We get :

```A(n) = A(n - 1) + n4
= A(n - 2) + n4  + (n-1)4
= A(n - 3) + n4  + (n-1)4 + (n-2)4
.
.
.
= A(1) + 16 + 81... + (n-1)4 + n4

A(n) = 1 + 16 + 81 +... + (n-1)4 + n4
= n(n + 1)(6n3 + 9n2 + n - 1) / 30

i.e A(n) is sum of 4th powers of First n natural numbers.
```

Below is the implementation of the above approach:

## C++

 `// CPP program to find the n-th ` `// term in series ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find nth term ` `int` `sumOfSeries(``int` `n) ` `{ ` `    ``return` `n * (n + 1) * (6 * n * n * n ` `                 ``+ 9 * n * n + n - 1) / 30; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << sumOfSeries(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the n-th ` `// term in series ` `import` `java.io.*; ` ` `  `class` `Series { ` ` `  `    ``// Function to find nth term ` `    ``static` `int` `sumOfSeries(``int` `n) ` `    ``{ ` `        ``return` `n * (n + ``1``) * (``6` `* n * n * n  ` `                    ``+ ``9` `* n * n + n - ``1``) / ``30``; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(sumOfSeries(n)); ` `    ``} ` `} `

## Python

 `# Python program to find the Nth  ` `# term in series ` `  `  `# Function to print nth term  ` `# of series  ` `def` `sumOfSeries(n): ` `    ``return` `n ``*` `(n ``+` `1``) ``*` `(``6` `*` `n ``*` `n ``*` `n  ` `                 ``+` `9` `*` `n ``*` `n ``+` `n ``-` `1``)``/` `30` `      `  `# Driver code  ` `n ``=` `4` `print` `sumOfSeries(n) `

## C#

 `// C# program to find the n-th ` `// term in series ` `using` `System; ` `class` `Series { ` ` `  `    ``// Function to find nth term ` `    ``static` `int` `sumOfSeries(``int` `n) ` `    ``{ ` `        ``return` `n * (n + 1) * (6 * n * n * n  ` `                  ``+ 9 * n * n + n - 1) / 30; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4; ` `        ``Console.WriteLine(sumOfSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67 `

## PHP

 ` `

Output:

```354
```

Time Complexity : O(1).

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