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Given a positive integer n, the task is to find the sum of binomial coefficient i.e

ⁿC₀ + ⁿC₁ + ⁿC₂ + ……. + ⁿC_n-1 + ⁿC_n

Examples:

Input : n = 4
Output : 16
⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄
= 1 + 4 + 6 + 4 + 1
= 16

Input : n = 5
Output : 32

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Brute Force):
The idea is to evaluate each binomial coefficient term i.e ⁿC_r, where 0 <= r <= n and calculate the sum of all the terms.

Below is the implementation of this approach:

C++

// CPP Program to find the sum of Binomial 
// Coefficient. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns value of Binomial Coefficient Sum 
int binomialCoeffSum(int n) 
{ 
    int C[n + 1][n + 1]; 
  
    // Calculate value of Binomial Coefficient 
    // in bottom up manner 
    for (int i = 0; i <= n; i++) { 
        for (int j = 0; j <= min(i, n); j++) { 
            // Base Cases 
            if (j == 0 || j == i) 
                C[i][j] = 1; 
  
            // Calculate value using previously 
            // stored values 
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; 
        } 
    } 
  
    // Calculating the sum. 
    int sum = 0; 
    for (int i = 0; i <= n; i++) 
        sum += C[n][i]; 
  
    return sum; 
} 
  
/* Driver program to test above function*/
int main() 
{ 
    int n = 4; 
    printf("%d", binomialCoeffSum(n)); 
    return 0; 
} 

Java

// Java Program to find the sum  
// of Binomial Coefficient. 
  
class GFG { 
      
    // Returns value of Binomial  
    // Coefficient Sum 
    static int binomialCoeffSum(int n) 
    { 
        int C[][] = new int[n + 1][n + 1]; 
      
        // Calculate value of Binomial  
        // Coefficient in bottom up manner 
        for (int i = 0; i <= n; i++) 
        { 
            for (int j = 0; j <= Math.min(i, n); j++) 
            { 
                // Base Cases 
                if (j == 0 || j == i) 
                    C[i][j] = 1; 
      
                // Calculate value using previously 
                // stored values 
                else
                    C[i][j] = C[i - 1][j - 1] + 
                              C[i - 1][j]; 
              
                  
            } 
        } 
      
        // Calculating the sum. 
        int sum = 0; 
        for (int i = 0; i <= n; i++) 
            sum += C[n][i]; 
      
        return sum; 
    } 
      
    /* Driver program to test above function*/
    public static void main(String[] args) 
    { 
        int n = 4; 
        System.out.println(binomialCoeffSum(n)); 
    } 
} 
  
// This code is contributed by prerna saini. 

Python3

# Python  Program to find the sum  
# of Binomial Coefficient. 
   
import math     
   
# Returns value of Binomial  
# Coefficient Sum 
def binomialCoeffSum( n): 
      
        C = [[0]*(n+2) for i in range(0,n+2)] 
       
        # Calculate value of Binomial  
        # Coefficient in bottom up manner 
        for i in range(0,n+1): 
            for j in range(0, min(i, n)+1): 
              
                # Base Cases 
                if (j == 0 or j == i): 
                    C[i][j] = 1
       
                # Calculate value using previously 
                # stored values 
                else: 
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j] 
       
        # Calculating the sum. 
        sum = 0
        for i in range(0,n+1): 
            sum += C[n][i] 
       
        return sum
      
       
# Driver program to test above function 
n = 4
print(binomialCoeffSum(n)) 
  
# This code is contributed by Gitanjali. 

C#

// C# program to find the sum 
// of Binomial Coefficient. 
using System; 
  
class GFG { 
  
    // Returns value of Binomial 
    // Coefficient Sum 
    static int binomialCoeffSum(int n) 
    { 
        int[, ] C = new int[n + 1, n + 1]; 
  
        // Calculate value of Binomial 
        // Coefficient in bottom up manner 
        for (int i = 0; i <= n; i++)  
        { 
            for (int j = 0; j <= Math.Min(i, n); j++)  
            { 
                // Base Cases 
                if (j == 0 || j == i) 
                    C[i, j] = 1; 
  
                // Calculate value using previously 
                // stored values 
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j]; 
            } 
        } 
  
        // Calculating the sum. 
        int sum = 0; 
        for (int i = 0; i <= n; i++) 
            sum += C[n, i]; 
  
        return sum; 
    } 
  
    /* Driver program to test above function*/
    public static void Main() 
    { 
        int n = 4; 
        Console.WriteLine(binomialCoeffSum(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP Program to find the  
// sum of Binomial Coefficient. 
// Returns value of Binomial  
// Coefficient Sum 
  
function binomialCoeffSum($n) 
{ 
    $C[$n + 1][$n + 1] = array(0); 
  
    // Calculate value of  
    // Binomial Coefficient 
    // in bottom up manner 
    for ($i = 0; $i <= $n; $i++)  
    { 
        for ($j = 0; 
             $j <= min($i, $n); $j++) 
        { 
            // Base Cases 
            if ($j == 0 || $j == $i) 
                $C[$i][$j] = 1; 
  
            // Calculate value  
            // using previously 
            // stored values 
            else
                $C[$i][$j] = $C[$i - 1][$j - 1] +  
                             $C[$i - 1][$j]; 
        } 
    } 
  
    // Calculating the sum. 
    $sum = 0; 
    for ($i = 0; $i <= $n; $i++) 
        $sum += $C[$n][$i]; 
  
    return $sum; 
} 
  
// Driver Code 
$n = 4; 
echo binomialCoeffSum($n); 
  
// This code is contributed by ajit 
?> 

Output:

Method 2 (Using Formula):

This can be proved in 2 ways.
First Proof: Using Principle of induction.

For basic step, n = 0
LHS = ⁰C₀ = (0!)/(0! * 0!) = 1/1 = 1.
RHS= 2⁰ = 1.
LHS = RHS

For induction step:
Let k be an integer such that k > 0 and for all r, 0 <= r <= k, where r belong to integers,
the formula stand true.
Therefore,
^kC₀ + ^kC₁ + ^kC₂ + ……. + ^kC_k-1 + ^kC_k = 2^k

Now, we have to prove for n = k + 1,
^k+1C₀ + ^k+1C₁ + ^k+1C₂ + ……. + ^k+1C_k + ^k+1C_k+1 = 2^k+1

LHS = ^k+1C₀ + ^k+1C₁ + ^k+1C₂ + ……. + ^k+1C_k + ^k+1C_k+1
(Using ⁿC₀ = 0 and ⁿ⁺¹C_r = ⁿC_r + ⁿC_r-1)
= 1 + ^kC₀ + ^kC₁ + ^kC₁ + ^kC₂ + …… + ^kC_k-1 + ^kC_k + 1
= ^kC₀ + ^kC₀ + ^kC₁ + ^kC₁ + …… + ^kC_k-1 + ^kC_k-1 + ^kC_k + ^kC_k
= 2 X ∑ ⁿC_r
= 2 X 2^k
= 2^k+1
= RHS

Second Proof: Using Binomial theorem expansion

Binomial expansion state,
(x + y)ⁿ = ⁿC₀ xⁿ y⁰ + ⁿC₁ x^n-1 y¹ + ⁿC₂ x^n-2 y² + ……… + ⁿC_n-1 x¹ y^n-1 + ⁿC_n x⁰ yⁿ

Put x = 1, y = 1
(1 + 1)ⁿ = ⁿC₀ 1ⁿ 1⁰ + ⁿC₁ x^n-1 1¹ + ⁿC₂ 1^n-2 1² + ……… + ⁿC_n-1 1¹ 1^n-1 + ⁿC_n 1⁰ 1ⁿ

2ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ……. + ⁿC_n-1 + ⁿC_n

Below is implementation of this approach:

C++

// CPP Program to find sum of Binomial 
// Coefficient. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns value of Binomial Coefficient Sum 
// which is 2 raised to power n. 
int binomialCoeffSum(int n) 
{ 
    return (1 << n); 
} 
  
/* Driver program to test above function*/
int main() 
{ 
    int n = 4; 
    printf("%d", binomialCoeffSum(n)); 
    return 0; 
} 

Java

// Java Program to find sum  
// of Binomial Coefficient. 
import java.io.*; 
  
class GFG 
{ 
    // Returns value of Binomial 
    // Coefficient Sum which is  
    // 2 raised to power n. 
    static int binomialCoeffSum(int n) 
    { 
        return (1 << n); 
    } 
  
    // Driver Code 
    public static void main (String[] args)  
    { 
        int n = 4; 
        System.out.println(binomialCoeffSum(n)); 
    } 
} 
  
// This code is contributed  
// by akt_mit. 

Python3

# Python  Program to find the sum  
# of Binomial Coefficient. 
   
import math      
# Returns value of Binomial  
# Coefficient Sum 
def binomialCoeffSum( n): 
      
    return (1 << n); 
  
# Driver program to test 
# above function 
n = 4
print(binomialCoeffSum(n)) 
  
# This code is contributed 
# by Gitanjali. 

C#

// C# Program to find sum of  
// Binomial Coefficient. 
using System; 
  
class GFG { 
  
    // Returns value of Binomial Coefficient Sum 
    // which is 2 raised to power n. 
    static int binomialCoeffSum(int n) 
    { 
        return (1 << n); 
    } 
  
    /* Driver program to test above function*/
    static public void Main() 
    { 
        int n = 4; 
        Console.WriteLine(binomialCoeffSum(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP Program to find sum  
// of Binomial Coefficient. 
  
// Returns value of Binomial  
// Coefficient Sum which is  
// 2 raised to power n. 
function binomialCoeffSum($n) 
{ 
    return (1 << $n); 
} 
  
// Driver Code 
$n = 4; 
echo binomialCoeffSum($n); 
  
// This code is contributed 
// by akt_mit 
?> 

Output:

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Sum of Binomial coefficients

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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