Given an integer N, the task is to find the Nth term of the following series:
0, 8, 64, 216, 512, 1000, 1728, . . .
Examples:
Input: N = 6
Output: 1000
Input: N = 5
Output: 512
Approach:
- Given series 0, 8, 64, 216, 512, 1000, 1728, … can also be written as 0 * (02), 2 * (22), 4 * (42), 6 * (62), 8 * (82), 10 * (102), …
- Observe that 0, 2, 4, 6, 10, … is in AP and the nth term of this series can be found using the formula term = a1 + (n – 1) * d where a1 is the first term, n is the term position and d is the common difference.
- To get the term in the original series, term = term * (term2) i.e. term3.
- Finally print the term.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the nth term of the given series long term( int n)
{ // Common difference
int d = 2;
// First term
int a1 = 0;
// nth term
int An = a1 + (n - 1) * d;
// nth term of the given series
An = pow (An, 3);
return An;
} // Driver code int main()
{ int n = 5;
cout << term(n);
return 0;
} |
Java
// Java implementation of the approach import java.util.*;
import java.lang.*;
import java.io.*;
public class GFG {
// Function to return the nth term of the given series
static int nthTerm( int n)
{
// Common difference and first term
int d = 2 , a1 = 0 ;
// nth term
int An = a1 + (n - 1 ) * d;
// nth term of the given series
return ( int )Math.pow(An, 3 );
}
// Driver code
public static void main(String[] args)
{
int n = 5 ;
System.out.println(nthTerm(n));
}
} |
Python3
# Python3 implementation of the approach # Function to return the nth term of the given series def term(n):
# Common difference
d = 2
# First term
a1 = 0
# nth term
An = a1 + (n - 1 ) * d
# nth term of the given series
An = An * * 3
return An;
# Driver code n = 5
print (term(n))
|
C#
// C# implementation of the approach using System;
public class GFG {
// Function to return the nth term of the given series
static int nthTerm( int n)
{
// Common difference and first term
int d = 2, a1 = 0;
// nth term
int An = a1 + (n - 1) * d;
// nth term of the given series
return ( int )Math.Pow(An, 3);
}
// Driver code
public static void Main()
{
int n = 5;
Console. WriteLine(nthTerm(n));
}
} // This code is contributed by Mutual singh. |
PHP
<?php // PHP implementation of the approach // Function to return the nth term of the given series function term( $n )
{ // Common difference
$d = 2;
// First term
$a1 = 0;
// nth term
$An = $a1 +( $n -1)* $d ;
// nth term of the given series
return pow( $An , 3);
} // Driver code $n = 5;
echo term( $n );
?> |
Javascript
<script> // javascript implementation of the approach // Function to return the nth term of the given series function term(n)
{ // Common difference
let d = 2;
// First term
let a1 = 0;
// nth term
An=a1+(n-1)*d;
// nth term of the given series
return Math.pow(An, 3);
} // Driver code let n = 5; document.write( term(n)); // This code is contributed by sravan kumar </script> |
Output:
512
Time Complexity: O(1)
Auxiliary Space: O(1)