If n be a positive integer and `( 1+x)^(n) = a_(0) + a_(1)x + a_(2)x^(2) +"………."+ a_(n)x^(n)`, then what is `a_(0) + a_(1) +a_(2)+"……….."+a_(n)` equal to ?

Options

A. 1
B. `2^(n)`
C. `2^(n-1)`
D. `2^(n+1)`

Answer

To solve the problem, we need to find the value of \( a_0 + a_1 + a_2 + \ldots + a_n \) from the expression \( (1 + x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The expression \( (1 + x)^n \) can be expanded using the Binomial Theorem, which states:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Here, \( a_k = \binom{n}{k} \).

2. **Finding the Sum of Coefficients**:
To find \( a_0 + a_1 + a_2 + \ldots + a_n \), we can evaluate the polynomial at \( x = 1 \):
\[ (1 + 1)^n = a_0 + a_1 + a_2 + \ldots + a_n \]

3. **Calculating the Left Side**:
Substitute \( x = 1 \) into the left-hand side:
\[ (1 + 1)^n = 2^n \]

4. **Equating Both Sides**:
From the above, we have:
\[ a_0 + a_1 + a_2 + \ldots + a_n = 2^n \]

5. **Final Answer**:
Therefore, the value of \( a_0 + a_1 + a_2 + \ldots + a_n \) is:
\[ \boxed{2^n} \]