Sum of all the numbers present at given level in Pascal’s triangle

Given a level L. The task is to find the sum of all the integers present at the given level in Pascal’s triangle .

A Pascal triangle with 6 levels is as shown below:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1



Examples:

Input: L = 3
Output: 4
1 + 2 + 1 = 4

Input: L = 2
Output:2

Approach: If we observe carefully the series of the sum of levels will go on like 1, 2, 4, 8, 16…., which is a GP series with a = 1 and r = 2.

Therefore, sum of Lth level is L’th term in the above series.

Lth term = 2L-1

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sum of numbers at 
// Lth level in Pascals Triangle
int sum(int h)
{
    return pow(2, h - 1);
}
  
// Driver Code
int main()
{
    int L = 3;
      
    cout << sum(L);
      
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG
{
      
    // Function to find sum of numbers at 
    // Lth level in Pascals Triangle 
    static int sum(int h) 
    
        return (int)Math.pow(2, h - 1); 
    
      
    // Driver Code 
    public static void main (String[] args)
    
        int L = 3
          
        System.out.println(sum(L)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the above approach
  
# Function to find sum of numbers at
# Lth level in Pascals Triangle
def summ(h):
    return pow(2, h - 1)
  
# Driver Code
L = 3
  
print(summ(L))
  
# This code is contributed by mohit kumar

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C#

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// C# implementation of the approach 
using System;
  
class GFG
{
      
    // Function to find sum of numbers at 
    // Lth level in Pascals Triangle 
    static int sum(int h) 
    
        return (int)Math.Pow(2, h - 1); 
    
      
    // Driver Code 
    public static void Main ()
    
        int L = 3; 
          
        Console.WriteLine(sum(L)); 
    
}
  
// This code is contributed by anuj_67..

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Output:

4

Time Complexity: O(1)



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