If the value of the sum `29(.^(30)C_(0))+28(.^(30)C_(1))+27(.^(30)C_(2))+…….+1(.^(30)C_(28))-(.^(30)C_(30))` is equal to `K.2^(32)`, then the value of K is equal to

Options

A. 7
B. 14
C. `(5)/(2)`
D. `(7)/(2)`

Answer

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

\[ S = 29 \cdot \binom{30}{0} + 28 \cdot \binom{30}{1} + 27 \cdot \binom{30}{2} + \ldots + 1 \cdot \binom{30}{28} - \binom{30}{30} \]

### Step 1: Rewrite the Sum
We can rewrite the sum \( S \) by including the term \( 0 \cdot \binom{30}{29} \):

\[ S = 29 \cdot \binom{30}{0} + 28 \cdot \binom{30}{1} + 27 \cdot \binom{30}{2} + \ldots + 1 \cdot \binom{30}{28} + 0 \cdot \binom{30}{29} - \binom{30}{30} \]

### Step 2: Express the Sum in Summation Notation
We can express this sum using summation notation:

\[ S = \sum_{k=0}^{30} (29 - k) \cdot \binom{30}{k} \]

### Step 3: Split the Summation
We can split the summation into two parts:

\[ S = \sum_{k=0}^{30} 29 \cdot \binom{30}{k} - \sum_{k=0}^{30} k \cdot \binom{30}{k} \]

### Step 4: Evaluate the First Summation
The first summation can be evaluated using the binomial theorem:

\[ \sum_{k=0}^{30} \binom{30}{k} = 2^{30} \]

Thus,

\[ \sum_{k=0}^{30} 29 \cdot \binom{30}{k} = 29 \cdot 2^{30} \]

### Step 5: Evaluate the Second Summation
The second summation can be evaluated using the identity \( k \cdot \binom{n}{k} = n \cdot \binom{n-1}{k-1} \):

\[ \sum_{k=0}^{30} k \cdot \binom{30}{k} = 30 \cdot \sum_{k=1}^{30} \binom{29}{k-1} = 30 \cdot 2^{29} \]

### Step 6: Combine the Results
Now we can combine the results:

\[ S = 29 \cdot 2^{30} - 30 \cdot 2^{29} \]

Factoring out \( 2^{29} \):

\[ S = 2^{29} (29 \cdot 2 - 30) = 2^{29} (58 - 30) = 2^{29} \cdot 28 \]

### Step 7: Relate to Given Expression
We know from the problem statement that:

\[ S = K \cdot 2^{32} \]

Setting the two expressions for \( S \) equal gives:

\[ 28 \cdot 2^{29} = K \cdot 2^{32} \]

### Step 8: Solve for K
Dividing both sides by \( 2^{29} \):

\[ 28 = K \cdot 2^{3} \]

Thus,

\[ K = \frac{28}{8} = 3.5 = \frac{7}{2} \]

### Final Answer
The value of \( K \) is:

\[ \boxed{\frac{7}{2}} \]