Given a number N, the task is to find the nth term of the series
5, 2, 19, 13, 41, 31, 71, 57….
It is given that value of n can range between 1 and 10000.
Examples:
Input: N = 4
Output:13Input: N = 15
Output:272
Approach: The problem looks very hard but approach is very simple. If the value of n is given as an odd number, the nth term will be ( ( n + 1 ) ^ 2 ) + n.
Otherwise, it will be ( ( n – 1 ) ^ 2 ) + n.
Implementation:
// C++ program to find nth term of // the series 5 2 13 41 #include<bits/stdc++.h> using namespace std;
// function to calculate nth term of the series int nthTermOfTheSeries( int n)
{ // to store the nth term of series
int nthTerm;
// if n is even number
if (n % 2 == 0)
nthTerm = pow (n - 1, 2) + n;
// if n is odd number
else
nthTerm = pow (n + 1, 2) + n;
// return nth term
return nthTerm;
} // Driver code int main()
{ int n;
n = 8;
cout << nthTermOfTheSeries(n) << endl;
n = 12;
cout << nthTermOfTheSeries(n) << endl;
n = 102;
cout << nthTermOfTheSeries(n) << endl;
n = 999;
cout << nthTermOfTheSeries(n) << endl;
n = 9999;
cout << nthTermOfTheSeries(n) << endl;
return 0;
} // This code is contributed // by Akanksha Rai |
// C program to find nth term of // the series 5 2 13 41 #include <math.h> #include <stdio.h> // function to calculate nth term of the series int nthTermOfTheSeries( int n)
{ // to store the nth term of series
int nthTerm;
// if n is even number
if (n % 2 == 0)
nthTerm = pow (n - 1, 2) + n;
// if n is odd number
else
nthTerm = pow (n + 1, 2) + n;
// return nth term
return nthTerm;
} // Driver code int main()
{ int n;
n = 8;
printf ( "%d\n" , nthTermOfTheSeries(n));
n = 12;
printf ( "%d\n" , nthTermOfTheSeries(n));
n = 102;
printf ( "%d\n" , nthTermOfTheSeries(n));
n = 999;
printf ( "%d\n" , nthTermOfTheSeries(n));
n = 9999;
printf ( "%d\n" , nthTermOfTheSeries(n));
return 0;
} |
// Java program to find nth term of the series 5 2 13 41 import java.lang.Math;
class GFG
{ // function to calculate nth term of the series static long nthTermOfTheSeries( int n)
{ // to store the nth term of series
long nthTerm;
// if n is even number
if (n % 2 == 0 )
nthTerm = ( long )Math.pow(n - 1 , 2 ) + n;
// if n is odd number
else
nthTerm = ( long )Math.pow(n + 1 , 2 ) + n;
// return nth term
return nthTerm;
} // Driver code public static void main(String[] args)
{ int n;
n = 8 ;
System.out.println( nthTermOfTheSeries(n));
n = 12 ;
System.out.println( nthTermOfTheSeries(n));
n = 102 ;
System.out.println( nthTermOfTheSeries(n));
n = 999 ;
System.out.println( nthTermOfTheSeries(n));
n = 9999 ;
System.out.println( nthTermOfTheSeries(n));
//This code is contributed by 29AjayKumar } } |
# Python3 program to find nth term # of the series 5 2 13 41 from math import pow
# function to calculate nth term # of the series def nthTermOfTheSeries(n):
# to store the nth term of series
# if n is even number
if (n % 2 = = 0 ):
nthTerm = pow (n - 1 , 2 ) + n
# if n is odd number
else :
nthTerm = pow (n + 1 , 2 ) + n
# return nth term
return nthTerm
# Driver code if __name__ = = '__main__' :
n = 8
print ( int (nthTermOfTheSeries(n)))
n = 12
print ( int (nthTermOfTheSeries(n)))
n = 102
print ( int (nthTermOfTheSeries(n)))
n = 999
print ( int (nthTermOfTheSeries(n)))
n = 9999
print ( int (nthTermOfTheSeries(n)))
# This code is contributed by # Shashank_Sharma |
// C# program to find nth term // of the series 5 2 13 41 using System;
class GFG
{ // function to calculate
// nth term of the series
static long nthTermOfTheSeries( int n)
{
// to store the nth term of series
long nthTerm;
// if n is even number
if (n % 2 == 0)
nthTerm = ( long )Math.Pow(n - 1, 2) + n;
// if n is odd number
else
nthTerm = ( long )Math.Pow(n + 1, 2) + n;
// return nth term
return nthTerm;
}
// Driver code
public static void Main()
{
int n;
n = 8;
Console.WriteLine(nthTermOfTheSeries(n));
n = 12;
Console.WriteLine( nthTermOfTheSeries(n));
n = 102;
Console.WriteLine( nthTermOfTheSeries(n));
n = 999;
Console.WriteLine( nthTermOfTheSeries(n));
n = 9999;
Console.WriteLine( nthTermOfTheSeries(n));
}
} // This code is contributed by Ryuga |
<?php // Php program to find nth term of // the series 5 2 13 41 // function to calculate nth term // of the series function nthTermOfTheSeries( $n )
{ // if n is even number
if ( $n % 2 == 0)
$nthTerm = pow( $n - 1, 2) + $n ;
// if n is odd number
else
$nthTerm = pow( $n + 1, 2) + $n ;
// return nth term
return $nthTerm ;
} // Driver code $n = 8;
echo nthTermOfTheSeries( $n ) . "\n" ;
$n = 12;
echo nthTermOfTheSeries( $n ) . "\n" ;
$n = 102;
echo nthTermOfTheSeries( $n ) . "\n" ;
$n = 999;
echo nthTermOfTheSeries( $n ) . "\n" ;
$n = 9999;
echo nthTermOfTheSeries( $n ) . "\n" ;
// This code is contributed by ita_c ?> |
<script> // Javascript program to find nth term of // the series 5 2 13 41 // function to calculate nth term of the series function nthTermOfTheSeries(n)
{ // to store the nth term of series
let nthTerm;
// if n is even number
if (n % 2 == 0)
nthTerm = Math.pow(n - 1, 2) + n;
// if n is odd number
else
nthTerm = Math.pow(n + 1, 2) + n;
// return nth term
return nthTerm;
} // Driver code let n; n = 8; document.write(nthTermOfTheSeries(n) + "<br>" );
n = 12; document.write(nthTermOfTheSeries(n) + "<br>" );
n = 102; document.write(nthTermOfTheSeries(n) + "<br>" );
n = 999; document.write(nthTermOfTheSeries(n) + "<br>" );
n = 9999; document.write(nthTermOfTheSeries(n) + "<br>" );
// This code is contributed by rishavmahato348. </script> |
57 133 10303 1000999 100009999
Time Complexity: O(1) as it is doing constant operations
Auxiliary Space: O(1) since no extra array is used space taken by this algorithm is constant