Given a number n, the task is to find the nth term of the series
1, 4, 15, 24, 45, 60, 91, 112, 153…..
where 0 < n < 100000000.
Examples:
Input: n = 10 Output: 180 Input: n = 5 Output: 45
Approach:
The idea is very simple but hard to recognise.
If n is odd, the nth term will be [ ( 2 * ( n^2 ) ) – n ].
If n is even, the nth term will be [ 2 * ( (n^2) – n ) ].
Implementation:
C++
#include <stdio.h> // function to calculate nth term of the series long long int nthTerm( long long int n)
{ // variable nth will store the nth term of series
long long int nth;
// if n is even
if (n % 2 == 0)
nth = 2 * ((n * n) - n);
// if n is odd
else
nth = (2 * n * n) - n;
// return nth term
return nth;
} // Driver code int main()
{ long long int n;
n = 5;
printf ( "%lld\n" , nthTerm(n));
n = 25;
printf ( "%lld\n" , nthTerm(n));
n = 25000000;
printf ( "%lld\n" , nthTerm(n));
n = 250000007;
printf ( "%lld\n" , nthTerm(n));
return 0;
} |
Java
// Java implementation of the above approach class GFG
{ // function to calculate nth
// term of the series
static long nthTerm( long n)
{
// variable nth will store the
// nth term of series
long nth;
// if n is even
if (n % 2 == 0 )
nth = 2 * ((n * n) - n);
// if n is odd
else
nth = ( 2 * n * n) - n;
// return nth term
return nth;
}
// Driver code
public static void main(String []args)
{
long n;
n = 5 ;
System.out.println(nthTerm(n));
n = 25 ;
System.out.println(nthTerm(n));
n = 25000000 ;
System.out.println(nthTerm(n));
n = 250000007 ;
System.out.println(nthTerm(n));
}
} // This code is contributed by Ryuga |
Python3
# function to calculate nth term of the series def nthTerm(n):
# variable nth will store the nth
# term of series
nth = 0
# if n is even
if (n % 2 = = 0 ):
nth = 2 * ((n * n) - n)
# if n is odd
else :
nth = ( 2 * n * n) - n
# return nth term
return nth
# Driver code n = 5
print (nthTerm(n))
n = 25
print (nthTerm(n))
n = 25000000
print (nthTerm(n))
n = 250000007
print (nthTerm(n))
# This code is contributed by # Mohit kumar 29 |
C#
// C# implementation of the above approach using System;
class GFG
{ // function to calculate nth
// term of the series
static long nthTerm( long n)
{
// variable nth will store the
// nth term of series
long nth;
// if n is even
if (n % 2 == 0)
nth = 2 * ((n * n) - n);
// if n is odd
else
nth = (2 * n * n) - n;
// return nth term
return nth;
}
// Driver code
public static void Main()
{
long n;
n = 5;
Console.WriteLine(nthTerm(n));
n = 25;
Console.WriteLine(nthTerm(n));
n = 25000000;
Console.WriteLine(nthTerm(n));
n = 250000007;
Console.WriteLine(nthTerm(n));
}
} // This code is contributed by chandan_jnu |
PHP
<?php // function to calculate nth term // of the series function nthTerm( $n )
{ // variable nth will store the
// nth term of series
$nth ;
// if n is even
if ( $n % 2 == 0)
$nth = 2 * (( $n * $n ) - $n );
// if n is odd
else
$nth = (2 * $n * $n ) - $n ;
// return nth term
return $nth ;
} // Driver code $n = 5;
echo nthTerm( $n ), "\n" ;
$n = 25;
echo nthTerm( $n ), "\n" ;
$n = 25000000;
echo nthTerm( $n ), "\n" ;
$n = 250000007;
echo nthTerm( $n ), "\n" ;
// This code is contributed by jit_t ?> |
Javascript
<script> // function to calculate nth term of the series function nthTerm( n)
{ // variable nth will store the nth term of series
let nth;
// if n is even
if (n % 2 == 0)
nth = 2 * ((n * n) - n);
// if n is odd
else
nth = (2 * n * n) - n;
// return nth term
return nth;
} // Driver code let n = 5;
document.write( nthTerm(n) + "<br/>" );
n = 25;
document.write( nthTerm(n) + "<br/>" );
n = 25000000;
document.write( nthTerm(n) + "<br/>" );
n = 250000007;
document.write( nthTerm(n) + "<br/>" );
// This code contributed by gauravrajput1 </script> |
Output:
45 1225 1249999950000000 125000006750000091
Time Complexity: O(1)
Auxiliary Space: O(1)