Given a number N, the task is to print the first N terms of the series:
Examples:
Input: N = 7
Output: 1, 9, 17, 33, 49, 73, 97
Input: N = 3
Output: 1, 9, 17
Approach: From the given series, find the formula for Nth term:
1st term = 1 2nd term = 9 = 2 * 4 + 1 3rd term = 17 = 2 * 9 - 1 4th term = 33 = 2 * 16 + 1 5th term = 49 = 2 * 25 - 1 6th term = 73 = 2 * 36 + 1 . . Nth term = (2 * N2 + (-1)N)
Therefore:
Nth term of the series
*** QuickLaTeX cannot compile formula: *** Error message: Error: Nothing to show, formula is empty
Then iterate over numbers in the range [1, N] to find all the terms using the above formula and print them.
Below is the implementation of the above approach:
CPP
// C++ implementation of the above approach #include "bits/stdc++.h" using namespace std;
// Function to print the series void printSeries( int N)
{ int ith_term = 0;
// Generate the ith term and
// print it
for ( int i = 1; i <= N; i++) {
ith_term = i % 2 == 0
? 2 * i * i + 1
: 2 * i * i - 1;
cout << ith_term << ", " ;
}
} // Driver Code int main()
{ int N = 7;
printSeries(N);
return 0;
} |
Java
// Java implementation of the above approach import java.util.*;
class GFG{
// Function to print the series static void printSeries( int N)
{ int ith_term = 0 ;
// Generate the ith term and
// print it
for ( int i = 1 ; i <= N; i++) {
ith_term = i % 2 == 0
? 2 * i * i + 1
: 2 * i * i - 1 ;
System.out.print(ith_term+ ", " );
}
} // Driver Code public static void main(String[] args)
{ int N = 7 ;
printSeries(N);
} } // This code is contributed by PrinciRaj1992 |
Python3
# Python implementation of the above approach # Function to print series def printSeries(N):
ith_term = 0 ;
# Generate the ith term and
# print
for i in range ( 1 ,N + 1 ):
ith_term = 0 ;
if (i % 2 = = 0 ):
ith_term = 2 * i * i + 1 ;
else :
ith_term = 2 * i * i - 1 ;
print (ith_term,end = ", " );
# Driver Code if __name__ = = '__main__' :
N = 7 ;
printSeries(N);
# This code is contributed by Princi Singh |
C#
// C# implementation of the above approach using System;
class GFG{
// Function to print the series static void printSeries( int N)
{ int ith_term = 0;
// Generate the ith term and
// print it
for ( int i = 1; i <= N; i++) {
ith_term = i % 2 == 0? 2 * i * i + 1:
2 * i * i - 1;
Console.Write(ith_term+ ", " );
}
} // Driver Code public static void Main()
{ int N = 7;
printSeries(N);
} } // This code is contributed by AbhiThakur |
Javascript
<script> // javascript implementation of the above approach // Function to print the series function printSeries( N)
{ let ith_term = 0;
// Generate the ith term and
// print it
for (let i = 1; i <= N; i++)
{
ith_term = i % 2 == 0
? 2 * i * i + 1
: 2 * i * i - 1;
document.write( ith_term + ", " );
}
} // Driver Code let N = 7;
printSeries(N);
// This code is contributed by gauravrajput1
</script> |
Output:
1, 9, 17, 33, 49, 73, 97,
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.