Given a number N. The task is to write a program to find the Nth term of the below series:
0, 5, 18, 39, 67, 105, 150, 203, …(N Terms)
Examples:
Input: N = 4
Output: 82
For N = 4
4th Term = ( 4 * 4 * 4 - 7 * 4 + 3)
= 39
Input: N = 10
Output: 332
Approach: The generalized Nth term of this series:
Below is the required implementation:
C++
// CPP program to find the N-th term of the series:// 0, 5, 18, 39, 67, 105, 150, 203, ...#include <iostream>#include <math.h>using namespace std;// calculate Nth term of seriesint nthTerm(int n){ return 4 * pow(n, 2) - 7 * n + 3;}// Driver codeint main(){ int N = 4; cout << nthTerm(N); return 0;} |
Java
// Java program to find the N-th term of the series: // 0, 5, 18, 39, 67, 105, 150, 203, import java.util.*;import java.lang.*;import java.io.*;class GFG{// calculate Nth term of series static int nthTerm(int n) { return 4 * (int)Math.pow(n, 2) - 7 * n + 3; } // Driver code public static void main(String args[]) { int N = 4; System.out.print(nthTerm(N)); } } |
Python 3
# Python 3 program to find the # N-th term of the series:# 0, 5, 18, 39, 67, 105, 150, 203, ...# calculate Nth term of seriesdef nthTerm(n): return 4 * pow(n, 2) - 7 * n + 3# Driver codeN = 4print(nthTerm(N))# This code is contributed by ash264 |
C#
// C# program to find the // N-th term of the series: // 0, 5, 18, 39, 67, 105, 150, 203, using System;class GFG{// calculate Nth term of series static int nthTerm(int n) { return 4 * (int)Math.Pow(n, 2) - 7 * n + 3; } // Driver code static public void Main (){ int N = 4; Console.Write(nthTerm(N)); } } // This code is contributed by Raj |
PHP
<?php// PHP program to find the // N-th term of the series:// 0, 5, 18, 39, 67, 105, 150, 203, ...// calculate Nth term of seriesfunction nthTerm($n){ return 4 * pow($n, 2) - 7 * $n + 3;}// Driver code$N = 4;echo nthTerm($N);// This code is contributed // by Akanksha Rai(Abby_akku)?> |
Javascript
<script>// JavaScript program to find the N-th term of the series:// 0, 5, 18, 39, 67, 105, 150, 203, ...// calculate Nth term of seriesfunction nthTerm( n){ return 4 * Math.pow(n, 2) - 7 * n + 3;}// driver code // declaration of number of terms let N = 4; document.write( nthTerm(N) );// This code contributed by gauravrajput1</script> |
Output:
39
Time Complexity: O(1)
Space Complexity: O(1) since using constant variables
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!
