Sum of Binomial coefficients

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

 

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: 
 

Output: 

16

Method 2 (Using Formula): 
 

👁 Image

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: 
 

Output: 
 

16