第N个括号中所有数字的总和

给定一个整数N和一个序列(1)，(3、5)，(7、9、11)，(13、15、17、19)，…..任务是找到其中所有数字的总和。^第N^个括号。
例子：

Input: N = 2 
Output: 8 
3 + 5 = 8
Input: N = 3 
Output: 27 
7 + 9 + 11 = 27 
 

方法：可以观察到，对于N = 1，2，3，…的序列将形成为1，8，27，64，125，216，343，…，其^第N^个项是N ³
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

 
// 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));
 

输出：

27

时间复杂度： O(1)