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++
// CPP program to find N-th term of the series: // 0, 5, 14, 27, 44 ... #include <iostream> #include <math.h> using namespace std;
// Calculate Nth term of series int nthTerm( int n)
{ return 2 * pow (n, 2) - n - 1;
} // Driver code int main()
{ int N = 4;
cout << nthTerm(N);
return 0;
} |
Java
// Java program to find N-th term of the series: // 0, 5, 14, 27, 44 ... import java.util.*;
class solution
{ // Calculate Nth term of series static int nthTerm( int n)
{ return 2 *( int )Math.pow(n, 2 ) - n - 1 ;
} // Driver code public static void main(String arr[])
{ int N = 4 ;
System.out.println(nthTerm(N));
} } //This code is contributed by Surendra_Gangwar |
Python 3
# Python 3 program to find # N-th term of the series: # 0, 5, 14, 27, 44 ... # Calculate Nth term of series def nthTerm(n):
return 2 * pow (n, 2 ) - n - 1
# Driver code if __name__ = = "__main__" :
N = 4
print (nthTerm(N))
# This code is contributed # by ChitraNayal |
C#
// C# program to find // N-th term of the series: // 0, 5, 14, 27, 44 ... using System;
class GFG
{ // Calculate Nth term of series static int nthTerm( int n)
{ return 2 * ( int )Math.Pow(n, 2) - n - 1;
} // Driver code static public void Main ()
{ int N = 4;
Console.Write(nthTerm(N));
} } // This code is contributed by Raj |
PHP
<?php // PHP program to find // N-th term of the series: // 0, 5, 14, 27, 44 ... // Calculate Nth term of series function nthTerm( $n )
{ return 2 * pow( $n , 2) - $n - 1;
} // Driver code $N = 4;
echo nthTerm( $N );
// This code is contributed // by Akanksha Rai(Abby_akku) ?> |
Javascript
<script> // JavaScript program to find N-th term of the series: // 0, 5, 14, 27, 44 ... // Calculate Nth term of series function nthTerm( n)
{ return 2 * Math.pow(n, 2) - n - 1;
} // Driver code let N = 4;
document.write( nthTerm(N) );
// This code contributed by aashish1995 </script> |
Output:
27
Time Complexity: O(1)
Auxiliary Space : O(1) because using constant variables
Note: Sum upto n terms of the above series(Sn) is: