Given a number N, the task is to find the Nth term in series 12, 35, 81, 173, 357, …
Example:
Input: N = 2 Output: 35 2nd term = (12*2) + 11 = 35 Input: N = 5 Output: 357 5th term = (12*(2^4))+11*((2^4)-1) = 357
Approach:
- Each and every number is obtained by multiplying the previous number by 2 and the addition of 11 to it.
- Since starting number is 12.
1st term = 12 2nd term = (12 * 2) / 11 = 35 3rd term = (35 * 2) / 11 = 81 4th term = (81 * 2) / 11 = 173 And, so on....
- In general, Nth number is obtained by formula:
Below is the implementation of the above approach:
C++
// C++ program to find the Nth term // in series 12, 35, 81, 173, 357, ... #include <bits/stdc++.h> using namespace std;
// Function to find Nth term int nthTerm( int N)
{ int nth = 0, first_term = 12;
// Nth term
nth = (first_term * ( pow (2, N - 1)))
+ 11 * (( pow (2, N - 1)) - 1);
return nth;
} // Driver Method int main()
{ int N = 5;
cout << nthTerm(N) << endl;
return 0;
} |
Java
// Java program to find the Nth term // in series 12, 35, 81, 173, 357, ... class GFG
{ // Function to find Nth term static int nthTerm( int N)
{ int nth = 0 , first_term = 12 ;
// Nth term
nth = ( int ) ((first_term * (Math.pow( 2 , N - 1 )))
+ 11 * ((Math.pow( 2 , N - 1 )) - 1 ));
return nth;
} // Driver code public static void main(String[] args)
{ int N = 5 ;
System.out.print(nthTerm(N) + "\n" );
} } // This code is contributed by Rajput-Ji |
Python3
# Python3 program to find the Nth term # in series 12, 35, 81, 173, 357, ... # Function to find Nth term def nthTerm(N) :
nth = 0 ; first_term = 12 ;
# Nth term
nth = (first_term * ( pow ( 2 , N - 1 ))) + \
11 * (( pow ( 2 , N - 1 )) - 1 );
return nth;
# Driver Method if __name__ = = "__main__" :
N = 5 ;
print (nthTerm(N)) ;
# This code is contributed by AnkitRai01 |
C#
// C# program to find the Nth term // in series 12, 35, 81, 173, 357, ... using System;
class GFG
{ // Function to find Nth term static int nthTerm( int N)
{ int nth = 0, first_term = 12;
// Nth term
nth = ( int ) ((first_term * (Math.Pow(2, N - 1)))
+ 11 * ((Math.Pow(2, N - 1)) - 1));
return nth;
} // Driver code public static void Main(String[] args)
{ int N = 5;
Console.Write(nthTerm(N) + "\n" );
} } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
// Function to find Nth term
function nthTerm(N)
{
let nth = 0, first_term = 12;
// Nth term
nth = (first_term * (Math.pow(2, N - 1)))
+ 11 * ((Math.pow(2, N - 1)) - 1);
return nth;
}
let N = 5;
document.write(nthTerm(N));
// This code is contributed by divyeshrabadiya07. </script> |
Output:
357
Time complexity: O(log N) for given input N because using inbuilt pow function
Auxiliary Space: O(1)