`1/2. ""^nC_0 + ""^nC_1 + 2. ""^nC_2 + 2^2. ""^nC_3 + …….+ 2^(n-1) . ""^nC_n = `

Options

A. `(3^(-n))/(2)`
B. `(4^(2n))/(2)`
C. `(4^n)/(2)`
D. `(3^n)/(2)`

Answer

To solve the given series

\[ \frac{1}{2} \binom{n}{0} + \binom{n}{1} + 2 \cdot \binom{n}{2} + 2^2 \cdot \binom{n}{3} + \ldots + 2^{n-1} \cdot \binom{n}{n} \]

we can follow these steps:

### Step 1: Factor out \(\frac{1}{2}\)

We start by factoring out \(\frac{1}{2}\) from the entire series:

\[ \frac{1}{2} \left( \binom{n}{0} + \binom{n}{1} \cdot 2 + \binom{n}{2} \cdot 2^2 + \ldots + \binom{n}{n} \cdot 2^{n-1} \right) \]

### Step 2: Rewrite the series

The series inside the parentheses can be rewritten as:

\[ \binom{n}{0} \cdot 2^0 + \binom{n}{1} \cdot 2^1 + \binom{n}{2} \cdot 2^2 + \ldots + \binom{n}{n} \cdot 2^{n} \]

This is the expansion of \((1 + 2)^n\) using the Binomial Theorem, which states that:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} \]

### Step 3: Apply the Binomial Theorem

In our case, let \(x = 2\) and \(y = 1\):

\[ (1 + 2)^n = 3^n \]

Thus, we have:

\[ \binom{n}{0} \cdot 2^0 + \binom{n}{1} \cdot 2^1 + \binom{n}{2} \cdot 2^2 + \ldots + \binom{n}{n} \cdot 2^{n} = 3^n \]

### Step 4: Substitute back into the equation

Now, substituting back into our equation gives:

\[ \frac{1}{2} \cdot 3^n \]

### Step 5: Final result

Thus, the final result of the series is:

\[ \frac{3^n}{2} \]

### Conclusion

The answer to the given series is:

\[ \frac{3^n}{2} \]