If `S=(1)/(2) ""^(10)C_(0) -""^(10)C_(1) +2""^(10)C_(2)-2^(2)" "^(10)C_(3)…+2^(9)" "^(10)C_(10)`, then S is equal to

Options

A. 0
B. `(1)/(2)`
C. `(1)/(2)3^(10)`
D. none

Answer

To solve the problem, we need to evaluate the expression:

\[ S = \frac{1}{2} \binom{10}{0} - \binom{10}{1} + 2 \binom{10}{2} - 2^2 \binom{10}{3} + \ldots + 2^9 \binom{10}{10} \]

### Step 1: Rewrite the expression for S

We can express \( S \) in a more manageable form. Notice that the terms alternate in sign and involve powers of 2 multiplied by binomial coefficients. We can rewrite \( S \) as:

\[ S = \sum_{k=0}^{10} (-1)^k 2^{k-1} \binom{10}{k} \]

### Step 2: Multiply S by 2

To find a relationship, we multiply \( S \) by 2:

\[ 2S = 2 \cdot \left( \frac{1}{2} \binom{10}{0} - \binom{10}{1} + 2 \binom{10}{2} - 2^2 \binom{10}{3} + \ldots + 2^9 \binom{10}{10} \right) \]

This gives:

\[ 2S = \binom{10}{0} - 2 \binom{10}{1} + 2^2 \binom{10}{2} - 2^3 \binom{10}{3} + \ldots + 2^{10} \binom{10}{10} \]

### Step 3: Recognize the binomial expansion

The expression for \( 2S \) can be recognized as the expansion of \( (1 - 2)^{10} \):

\[ (1 - 2)^{10} = (-1)^{10} = 1 \]

### Step 4: Set up the equation

Thus, we have:

\[ 2S = 1 \]

### Step 5: Solve for S

Now we can solve for \( S \):

\[ S = \frac{1}{2} \]

### Final Answer

Therefore, the value of \( S \) is:

\[ S = \frac{1}{2} \]

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