`C_0 + 2.C_1 + 4.C_2 + …….+C_n.2^n = 243` , then n =

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the equation \( C_0 + 2C_1 + 4C_2 + \ldots + C_n \cdot 2^n = 243 \), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Series**:
The given series can be expressed in terms of binomial coefficients. Specifically, we can write:
\[ C_0 + 2C_1 + 4C_2 + \ldots + C_n \cdot 2^n = \sum_{k=0}^{n} C_k \cdot 2^k \]
where \( C_k = \binom{n}{k} \).

2. **Applying the Binomial Theorem**:
According to the Binomial Theorem, we know that:
\[ (1 + x)^n = \sum_{k=0}^{n} C_k \cdot x^k \]
If we let \( x = 2 \), we have:
\[ (1 + 2)^n = \sum_{k=0}^{n} C_k \cdot 2^k \]
Thus, we can rewrite our series as:
\[ 3^n = \sum_{k=0}^{n} C_k \cdot 2^k \]

3. **Setting the Equation**:
From the problem statement, we have:
\[ 3^n = 243 \]

4. **Finding the Value of \( n \)**:
We know that \( 243 \) can be expressed as a power of \( 3 \):
\[ 243 = 3^5 \]
Therefore, we can equate the exponents:
\[ n = 5 \]

5. **Conclusion**:
The value of \( n \) is:
\[ \boxed{5} \]