Program to find the Nth term of the series 0, 5, 14, 27, 44, ……..
Given a number N. The task is to write a program to find the Nth term in the below series:
0, 5, 14, 27, 44 …(Nth Term)
Examples:
Input: N = 4
Output: 27
For N = 4,
Nth term = ( 2 * N * N - N - 1 )
= ( 2 * 4 * 4 - 4 - 1 )
= 27
Input: N = 10
Output: 188
Approach: The generalized Nth term of this series:
Nth Term: 2 * N * N - N - 1
Below is the required implementation:
C++
#include <iostream>
#include <math.h>
using namespace std;
int nthTerm( int n)
{
return 2 * pow (n, 2) - n - 1;
}
int main()
{
int N = 4;
cout << nthTerm(N);
return 0;
}
|
Java
import java.util.*;
class solution
{
static int nthTerm( int n)
{
return 2 *( int )Math.pow(n, 2 ) - n - 1 ;
}
public static void main(String arr[])
{
int N = 4 ;
System.out.println(nthTerm(N));
}
}
|
Python 3
def nthTerm(n):
return 2 * pow (n, 2 ) - n - 1
if __name__ = = "__main__" :
N = 4
print (nthTerm(N))
|
C#
using System;
class GFG
{
static int nthTerm( int n)
{
return 2 * ( int )Math.Pow(n, 2) - n - 1;
}
static public void Main ()
{
int N = 4;
Console.Write(nthTerm(N));
}
}
|
PHP
<?php
function nthTerm( $n )
{
return 2 * pow( $n , 2) - $n - 1;
}
$N = 4;
echo nthTerm( $N );
?>
|
Javascript
<script>
function nthTerm( n)
{
return 2 * Math.pow(n, 2) - n - 1;
}
let N = 4;
document.write( nthTerm(N) );
</script>
|
Time Complexity: O(1)
Auxiliary Space : O(1) because using constant variables
Note: Sum upto n terms of the above series(Sn) is:
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