Given the first two terms of the series as 1 and 6 and all the elements of the series are 2 less than the mean of the number preceding and succeeding it. The task is to print the nth term of the series.
First few terms of the series are:
1, 6, 15, 28, 45, 66, 91, …
Examples:
Input: N = 3
Output: 15
Input: N = 1
Output: 1
Approach: The given series represents odd positioned numbers in the triangular number series. Since the nth triangular number can easily be found by (n * (n + 1) / 2), so for finding the odd numbers we can replace n by (2 * n) – 1 as (2 * n) – 1 will always result in odd numbers i.e. the nth number of the given series will be ((2 * n) – 1) * n.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the nth term // of the given series int oddTriangularNumber( int N)
{ return (N * ((2 * N) - 1));
} // Driver code int main()
{ int N = 3;
cout << oddTriangularNumber(N);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to return the nth term // of the given series static int oddTriangularNumber( int N)
{ return (N * (( 2 * N) - 1 ));
} // Driver code public static void main(String[] args)
{ int N = 3 ;
System.out.println(oddTriangularNumber(N));
} } // This code contributed by Rajput-Ji |
# Python 3 implementation of the approach # Function to return the nth term # of the given series def oddTriangularNumber(N):
return (N * (( 2 * N) - 1 ))
# Driver code if __name__ = = '__main__' :
N = 3
print (oddTriangularNumber(N))
# This code is contributed by # Surendra_Gangwar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the nth term
// of the given series
static int oddTriangularNumber( int N)
{
return (N * ((2 * N) - 1));
}
// Driver code
public static void Main(String[] args)
{
int N = 3;
Console.WriteLine(oddTriangularNumber(N));
}
} /* This code contributed by PrinciRaj1992 */ |
<?php // PHP implementation of the approach // Function to return the nth term // of the given series function oddTriangularNumber( $N )
{ return ( $N * ((2 * $N ) - 1));
} // Driver code
$N = 3;
echo oddTriangularNumber( $N );
// This code is contributed by Ryuga
?> |
<script> // Javascript implementation of the approach // Function to return the nth term // of the given series function oddTriangularNumber(N)
{ return (N * ((2 * N) - 1));
} // Driver code let N = 3; document.write(oddTriangularNumber(N)); // This code is contributed by subham348. </script> |
15
Time Complexity: O(1)
Auxiliary Space: O(1)