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