# Find Nth number of the series 1, 6, 15, 28, 45, …..

Given the first two numbers of a series. The task is to find the Nth( N may be up to 10^18) number of that series.

Note: Every Element in the array is two less than the mean of the number preceding and succeeding of it. The answer can be very large so print answer under modulo 10^9+9.

Examples:

```Input: N = 3
Output: 15
(1 + 15)/2 - 2 = 6

Input: N = 4
Output: 28
(6 + 28)/2 - 2 = 15
```

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

Observation: According to the statement, the series formed will be 1, 6, 15, 28, 45….. So, the formula for Nth term will be:

```2*n*n - n
```

## C++

 `// CPP program to find Nth term of the series ` `#include ` `using` `namespace` `std; ` `#define mod 1000000009 ` ` `  `// function to return nth term of the series ` `int` `NthTerm(``long` `long` `n) ` `{ ` `    ``long` `long` `x = (2 * n * n) % mod; ` `    ``return` `(x - n + mod) % mod; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``long` `long` `N = 4; ` ` `  `    ``// function call ` `    ``cout << NthTerm(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find N-th ` `// term of the series: ` ` `  `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// function to return nth term of the series ` `    ``static` `long` `NthTerm(``long` `n) ` `    ``{ ` `        ``long` `x = (``2` `* n * n) % ``1000000009``; ` `        ``return` `(x - n + ``1000000009``) % ``1000000009``; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` ` `  `        ``// Taking  n as 6 ` `        ``long` `N = ``4``; ` ` `  `        ``// Printing the nth term ` `        ``System.out.println(NthTerm(N)); ` `    ``} ` `} `

## Python

 `# Python 3 program to find   ` `# N-th term of the series:   ` `   `  ` `  `# Function for calculating   ` `# Nth term of series   ` `def` `NthTerm(N) :   ` `   `  `    ``# return nth term ` `    ``x ``=` `(``2` `*` `N``*``N)``%` `1000000009` `    ``return` `((x ``-` `N ``+` `1000000009``)``%` `1000000009``)   ` `   `  `# Driver code   ` `if` `__name__ ``=``=` `"__main__"` `:   ` `       `  `    ``N ``=` `4` `   `  `    ``# Function Calling   ` `    ``print``(NthTerm(N)) `

## C#

 `// C# program to find N-th ` `// term of the series: ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// function to return nth ` `// term of the series ` `static` `long` `NthTerm(``long` `n) ` `{ ` `    ``long` `x = (2 * n * n) % 1000000009; ` `    ``return` `(x - n + 1000000009) % ` `                    ``1000000009; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` `  `    ``// Taking n as 6 ` `    ``long` `N = 4; ` ` `  `    ``// Printing the nth term ` `    ``Console.WriteLine(NthTerm(N)); ` `} ` `} ` ` `  `// This code is contributed ` `// by inder_verma `

## PHP

 ` `

Output:

```28
```

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Improved By : inderDuMCA