Program to find the Nth term of the series 0, 14, 40, 78, 124, …
Given a number N. The task is to write a program to find the Nth term of the below series:
0, 14, 40, 78, 124 …(N Terms)
Examples:
Input: N = 4
Output: 78
For N = 4
Sum(upto 4 terms) = ( 6 * n * n - 4 * n - 2)
= ( 6 * 4 * 4 - 4 * 4 - 2)
= 78
Input: N = 10
Output: 557
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 6 * pow (n, 2) - 4 * n - 2;
}
int main()
{
int N = 4;
cout << nthTerm(N);
return 0;
}
|
Java
import java.util.*;
class solution
{
static int nthTerm( int n)
{
return 6 * ( int )Math.pow(n, 2 ) - 4 * n - 2 ;
}
public static void main(String arr[])
{
int N = 4 ;
System.out.println(nthTerm(N));
}
}
|
Python3
def nthTerm(n):
return int ( 6 * pow (n, 2 ) - 4 * n - 2 )
N = 4
print (nthTerm(N))
|
C#
using System;
class GFG
{
static int nthTerm( int n)
{
return 6 * ( int )Math.Pow(n, 2) -
4 * n - 2;
}
public static void Main()
{
int N = 4;
Console.WriteLine(nthTerm(N));
}
}
|
PHP
<?php
function nthTerm( $n )
{
return 6 * pow( $n , 2) - 4 * $n - 2;
}
$N = 4;
echo nthTerm( $N );
|
Javascript
<script>
function nthTerm( n)
{
return 6 * Math.pow(n, 2) - 4 * n - 2;
}
let N = 4;
document.write( nthTerm(N) );
</script>
|
Time Complexity: O(logN) because using inbuilt pow function
Space Complexity: O(1) since using constant variables
Note: Sum upto n terms of the above series(Sn) is:
Last Updated :
07 Aug, 2022
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