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 seriesint NthTerm(long long n){ long long x = (2 * n * n) % mod; return (x - n + mod) % mod;}// Driver codeint main(){ long long N = 4; // function call cout << NthTerm(N); return 0;} |
C
// C program to find Nth term of the series#include <stdio.h>#define mod 1000000009// function to return nth term of the seriesint NthTerm(long long n){ long long x = (2 * n * n) % mod; return (x - n + mod) % mod;}// Driver codeint main(){ long long N = 4; // function call printf("%d",NthTerm(N)); return 0;}// This code is contributed by kothavvsaakash. |
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 seriesstatic long NthTerm(long n){ long x = (2 * n * n) % 1000000009; return (x - n + 1000000009) % 1000000009;}// Driver Codepublic 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$mod = 1000000009;// function to return nth // term of the seriesfunction NthTerm($n){ global $mod; $x = (2 * $n * $n) % $mod; return ($x - $n + $mod) % $mod;}// Driver code$N = 4;// function callecho 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
Time Complexity: O(1)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!