Program to find the Nth term of series 0, 4, 14, 30, 51, 80, 114, 154, 200, …
Given a number N. The task is to write a program to find the Nth term of the below series:
0, 4, 14, 30, 51, 80, 114, 154, 200, …(N Terms)
Examples:
Input: N = 4
Output: 82
For N = 4
4th Term = ( 4 * 4 - 2 * 4 + 2)
= 10
Input: N = 10
Output: 122
Approach: The generalized Nth term of this series:
Below is the required implementation:
C++
#include <iostream>
#include <math.h>
using namespace std;
int nthTerm( int n)
{
return pow (n, 2) - 2 * n + 2;
}
int main()
{
int N = 4;
cout << nthTerm(N);
return 0;
}
|
Java
import java.util.*;
class solution
{
static int nthTerm( int n)
{
return ( int )Math.pow(n, 2 ) - 2 * n + 2 ;
}
public static void main(String arr[])
{
int N = 4 ;
System.out.println(nthTerm(N));
}
}
|
Python3
from math import *
def nthTerm(n) :
return pow (n, 2 ) - 2 * n + 2
if __name__ = = "__main__" :
N = 4
print (nthTerm(N))
|
C#
using System;
class GFG
{
public static int nthTerm( int n)
{
return ( int )Math.Pow(n, 2) -
2 * n + 2;
}
public static void Main( string [] arr)
{
int N = 4;
Console.WriteLine(nthTerm(N));
}
}
|
PHP
<?php
function nthTerm( $n )
{
return pow( $n , 2) - 2 * $n + 2;
}
$N = 4;
echo nthTerm( $N );
?>
|
Javascript
<script>
function nthTerm( n)
{
return Math.pow(n, 2) - 2 * n + 2;
}
let N = 4;
document.write( nthTerm(N) );
</script>
|
Time Complexity: O(1), since there is no loop or recursion.
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 geeks!
Last Updated :
28 Aug, 2022
Like Article
Save Article