Given a number N, the task is to find the nth term of the series
5, 2, 19, 13, 41, 31, 71, 57….
It is given that value of n can range between 1 and 10000.
Examples:
Input: N = 4
Output:13Input: N = 15
Output:272
Approach: The problem looks very hard but approach is very simple. If the value of n is given as an odd number, the nth term will be ( ( n + 1 ) ^ 2 ) + n.
Otherwise, it will be ( ( n – 1 ) ^ 2 ) + n.
Implementation:
C++
// C++ program to find nth term of // the series 5 2 13 41#include<bits/stdc++.h>using namespace std;// function to calculate nth term of the seriesint nthTermOfTheSeries(int n){ // to store the nth term of series int nthTerm; // if n is even number if (n % 2 == 0) nthTerm = pow(n - 1, 2) + n; // if n is odd number else nthTerm = pow(n + 1, 2) + n; // return nth term return nthTerm;}// Driver codeint main(){ int n; n = 8; cout << nthTermOfTheSeries(n) << endl; n = 12; cout << nthTermOfTheSeries(n) << endl; n = 102; cout << nthTermOfTheSeries(n) << endl; n = 999; cout << nthTermOfTheSeries(n) << endl; n = 9999; cout << nthTermOfTheSeries(n) << endl; return 0;}// This code is contributed// by Akanksha Rai |
C
// C program to find nth term of // the series 5 2 13 41#include <math.h>#include <stdio.h>// function to calculate nth term of the seriesint nthTermOfTheSeries(int n){ // to store the nth term of series int nthTerm; // if n is even number if (n % 2 == 0) nthTerm = pow(n - 1, 2) + n; // if n is odd number else nthTerm = pow(n + 1, 2) + n; // return nth term return nthTerm;}// Driver codeint main(){ int n; n = 8; printf("%d\n", nthTermOfTheSeries(n)); n = 12; printf("%d\n", nthTermOfTheSeries(n)); n = 102; printf("%d\n", nthTermOfTheSeries(n)); n = 999; printf("%d\n", nthTermOfTheSeries(n)); n = 9999; printf("%d\n", nthTermOfTheSeries(n)); return 0;} |
Java
// Java program to find nth term of the series 5 2 13 41import java.lang.Math;class GFG{// function to calculate nth term of the seriesstatic long nthTermOfTheSeries(int n){ // to store the nth term of series long nthTerm; // if n is even number if (n % 2 == 0) nthTerm = (long)Math.pow(n - 1, 2) + n; // if n is odd number else nthTerm = (long)Math.pow(n + 1, 2) + n; // return nth term return nthTerm;}// Driver codepublic static void main(String[] args){ int n; n = 8; System.out.println( nthTermOfTheSeries(n)); n = 12; System.out.println( nthTermOfTheSeries(n)); n = 102; System.out.println( nthTermOfTheSeries(n)); n = 999; System.out.println( nthTermOfTheSeries(n)); n = 9999; System.out.println( nthTermOfTheSeries(n));//This code is contributed by 29AjayKumar}} |
Python3
# Python3 program to find nth term # of the series 5 2 13 41from math import pow# function to calculate nth term # of the seriesdef nthTermOfTheSeries(n): # to store the nth term of series # if n is even number if (n % 2 == 0): nthTerm = pow(n - 1, 2) + n # if n is odd number else: nthTerm = pow(n + 1, 2) + n # return nth term return nthTerm# Driver codeif __name__ == '__main__': n = 8 print(int(nthTermOfTheSeries(n))) n = 12 print(int(nthTermOfTheSeries(n))) n = 102 print(int(nthTermOfTheSeries(n))) n = 999 print(int(nthTermOfTheSeries(n))) n = 9999 print(int(nthTermOfTheSeries(n)))# This code is contributed by# Shashank_Sharma |
C#
// C# program to find nth term // of the series 5 2 13 41 using System;class GFG { // function to calculate // nth term of the series static long nthTermOfTheSeries(int n) { // to store the nth term of series long nthTerm; // if n is even number if (n % 2 == 0) nthTerm = (long)Math.Pow(n - 1, 2) + n; // if n is odd number else nthTerm = (long)Math.Pow(n + 1, 2) + n; // return nth term return nthTerm; } // Driver code public static void Main() { int n; n = 8; Console.WriteLine(nthTermOfTheSeries(n)); n = 12; Console.WriteLine( nthTermOfTheSeries(n)); n = 102; Console.WriteLine( nthTermOfTheSeries(n)); n = 999; Console.WriteLine( nthTermOfTheSeries(n)); n = 9999; Console.WriteLine( nthTermOfTheSeries(n)); }} // This code is contributed by Ryuga |
PHP
<?php // Php program to find nth term of // the series 5 2 13 41// function to calculate nth term // of the seriesfunction nthTermOfTheSeries($n){ // if n is even number if ($n % 2 == 0) $nthTerm = pow($n - 1, 2) + $n; // if n is odd number else $nthTerm = pow($n + 1, 2) + $n; // return nth term return $nthTerm;}// Driver code$n = 8;echo nthTermOfTheSeries($n) . "\n";$n = 12;echo nthTermOfTheSeries($n) . "\n";$n = 102;echo nthTermOfTheSeries($n) . "\n";$n = 999;echo nthTermOfTheSeries($n) . "\n";$n = 9999;echo nthTermOfTheSeries($n) . "\n";// This code is contributed by ita_c?> |
Javascript
<script>// Javascript program to find nth term of // the series 5 2 13 41// function to calculate nth term of the seriesfunction nthTermOfTheSeries(n){ // to store the nth term of series let nthTerm; // if n is even number if (n % 2 == 0) nthTerm = Math.pow(n - 1, 2) + n; // if n is odd number else nthTerm = Math.pow(n + 1, 2) + n; // return nth term return nthTerm;}// Driver codelet n;n = 8;document.write(nthTermOfTheSeries(n) + "<br>");n = 12;document.write(nthTermOfTheSeries(n) + "<br>");n = 102;document.write(nthTermOfTheSeries(n) + "<br>");n = 999;document.write(nthTermOfTheSeries(n) + "<br>");n = 9999;document.write(nthTermOfTheSeries(n) + "<br>");// This code is contributed by rishavmahato348.</script> |
Output:
57 133 10303 1000999 100009999
Time Complexity: O(1) as it is doing constant operations
Auxiliary Space: O(1) since no extra array is used space taken by this algorithm is constant
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