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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