If the value of the sum `29(.^(30)C_(0))+28(.^(30)C_(1))+27(.^(30)C_(2))+…….+1(.^(30)C_(28))+0.(.^(30)C_(29))-(.^(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:

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

We can rewrite this sum in a more manageable form. Notice that the coefficients (29, 28, ..., 1, 0) can be expressed as \(29 - k\) where \(k\) ranges from 0 to 29. Thus, we can express the sum as:

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

This can be split into two separate sums:

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

Now, we know that:

1. The sum of the binomial coefficients is given by:

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

For \(n = 30\):

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

Thus,

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

2. The sum involving \(k \cdot \binom{n}{k}\) can be evaluated using the identity:

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

For \(n = 30\):

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

Thus,

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

Now substituting back into our expression for \(S\):

\[ S = 29(2^{30} - 1) - (30 \cdot 2^{29} - 30) - 1 \]

Simplifying this gives:

\[ S = 29 \cdot 2^{30} - 29 - 30 \cdot 2^{29} + 30 - 1 \]

Combining like terms:

\[ S = 29 \cdot 2^{30} - 30 \cdot 2^{29} + 0 \]

Notice that \(30 \cdot 2^{29} = 15 \cdot 2^{30}\), thus:

\[ S = 29 \cdot 2^{30} - 15 \cdot 2^{30} = (29 - 15) \cdot 2^{30} = 14 \cdot 2^{30} \]

Now, we are given that \(S = K \cdot 2^{32}\). Therefore, we equate:

\[ 14 \cdot 2^{30} = K \cdot 2^{32} \]

Dividing both sides by \(2^{30}\):

\[ 14 = K \cdot 2^2 \]

Thus,

\[ K = \frac{14}{4} = 3.5 \]

So, the value of \(K\) is:

\[ \boxed{3.5} \]