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

  • Last Updated : 28 Apr, 2021

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

 

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 <bits/stdc++.h>
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




<?php
// PHP program to find Nth
// term of the series
 
$mod = 1000000009;
 
// function to return nth
// term of the series
function NthTerm($n)
{
    global $mod;
    $x = (2 * $n * $n) % $mod;
    return ($x - $n + $mod) % $mod;
}
 
// Driver code
$N = 4;
 
// function call
echo NthTerm($N);
 
// This code is contributed
// by inder_verma
?>

Javascript




<script>
 
// Javascript program to find N-th
// term of the series:
 
    // function to return nth term of the series
    function NthTerm(n) {
        var x = (2 * n * n) % 1000000009;
        return (x - n + 1000000009) % 1000000009;
    }
 
    // Driver Code
     
 
        // Taking n as 6
        var N = 4;
 
        // Printing the nth term
        document.write(NthTerm(N));
 
// This code contributed by gauravrajput1
 
</script>
Output: 
28

 




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