Given an integer N and a sequence (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), ….. the task is to find the sum of all the numbers in Nth parenthesis.
Examples:
Input: N = 2
Output: 8
3 + 5 = 8
Input: N = 3
Output: 27
7 + 9 + 11 = 27
Approach: It can be observed that for the values of N = 1, 2, 3, … a series will be formed as 1, 8, 27, 64, 125, 216, 343, … whose Nth term is N3
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the sum of the // numbers in the nth parenthesis int findSum( int n)
{ return pow (n, 3);
} // Driver code int main()
{ int n = 3;
cout << findSum(n);
return 0;
} |
Java
// Java implementation of the approach class GFG
{ // Function to return the sum of the // numbers in the nth parenthesis static int findSum( int n)
{ return ( int )Math.pow(n, 3 );
} // Driver code public static void main(String[] args)
{ int n = 3 ;
System.out.println(findSum(n));
} } // This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approach # Function to return the sum of the # numbers in the nth parenthesis def findSum(n) :
return n * * 3 ;
# Driver code if __name__ = = "__main__" :
n = 3 ;
print (findSum(n));
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the sum of the // numbers in the nth parenthesis static int findSum( int n)
{ return ( int )Math.Pow(n, 3);
} // Driver code public static void Main(String[] args)
{ int n = 3;
Console.WriteLine(findSum(n));
} } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach // Function to return the sum of the // numbers in the nth parenthesis function findSum(n)
{ return Math.pow(n, 3);
} // Driver code var n = 3;
document.write(findSum(n)); </script> |
Output:
27
Time Complexity: O(1)
Auxiliary Space: O(1)
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