The value of the expression `log_(2)(1+(1)/(2) sum_(k=1)^(11) ""^(12)C_(k))`:

Options

A. 11
B. 12
C. 13
D. 14

Answer

To solve the expression \( \log_2 \left( 1 + \frac{1}{2} \sum_{k=1}^{11} \binom{12}{k} \right) \), we will follow these steps:

### Step 1: Evaluate the summation
We need to evaluate the summation \( \sum_{k=1}^{11} \binom{12}{k} \). We know from the binomial theorem that:

\[ (1 + x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k \]

For \( n = 12 \) and \( x = 1 \):

\[ (1 + 1)^{12} = 2^{12} = \sum_{k=0}^{12} \binom{12}{k} \]

This gives us:

\[ \sum_{k=0}^{12} \binom{12}{k} = 4096 \]

### Step 2: Separate the terms
The summation can be separated into two parts:

\[ \sum_{k=0}^{12} \binom{12}{k} = \binom{12}{0} + \sum_{k=1}^{11} \binom{12}{k} + \binom{12}{12} \]

Where \( \binom{12}{0} = 1 \) and \( \binom{12}{12} = 1 \). Therefore:

\[ 4096 = 1 + \sum_{k=1}^{11} \binom{12}{k} + 1 \]

This simplifies to:

\[ \sum_{k=1}^{11} \binom{12}{k} = 4096 - 2 = 4094 \]

### Step 3: Substitute back into the expression
Now substitute this back into the original expression:

\[ 1 + \frac{1}{2} \sum_{k=1}^{11} \binom{12}{k} = 1 + \frac{1}{2} \cdot 4094 = 1 + 2047 = 2048 \]

### Step 4: Evaluate the logarithm
Now we need to evaluate:

\[ \log_2(2048) \]

Since \( 2048 = 2^{11} \), we have:

\[ \log_2(2048) = \log_2(2^{11}) = 11 \]

### Final Answer
Thus, the value of the expression \( \log_2 \left( 1 + \frac{1}{2} \sum_{k=1}^{11} \binom{12}{k} \right) \) is:

\[ \boxed{11} \]

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