Let `a=3^(1//224)+1` and for all `n ge 3`,
let `f(n)=""^(n)C_(0)a^(n-1)-""^(n)C_(1)a^(n-2)+""^(n)C_(2)a^(n-3)+...+(_1)^(n-1)*""^(n)C_(n-1)*a^(0).`
If the value of f(2016)+f(2017)=`3^(k)`, the value of K is

Options

A. 6
B. 8
C. 9
D. 10

Answer

To solve the problem, we need to evaluate the function \( f(n) \) defined as:

\[ f(n) = \binom{n}{0} a^{n-1} - \binom{n}{1} a^{n-2} + \binom{n}{2} a^{n-3} - \ldots + (-1)^{n-1} \binom{n}{n-1} a^0 \]

where \( a = 3^{1/224} + 1 \).

### Step 1: Rewrite \( f(n) \)

We can rewrite \( f(n) \) using the binomial theorem. We can factor out \( a \) from each term:

\[ f(n) = \sum_{k=0}^{n} (-1)^k \binom{n}{k} a^{n-k-1} \]

This can be simplified to:

\[ f(n) = \frac{1}{a} \left( (a - 1)^n - (-1)^n \right) \]

### Step 2: Calculate \( f(2016) \) and \( f(2017) \)

Now we need to find \( f(2016) \) and \( f(2017) \):

\[ f(2016) = \frac{1}{a} \left( (a - 1)^{2016} - (-1)^{2016} \right) \]

Since \( (-1)^{2016} = 1 \):

\[ f(2016) = \frac{1}{a} \left( (a - 1)^{2016} - 1 \right) \]

Similarly, for \( f(2017) \):

\[ f(2017) = \frac{1}{a} \left( (a - 1)^{2017} - (-1)^{2017} \right) \]

Since \( (-1)^{2017} = -1 \):

\[ f(2017) = \frac{1}{a} \left( (a - 1)^{2017} + 1 \right) \]

### Step 3: Add \( f(2016) \) and \( f(2017) \)

Now we add \( f(2016) \) and \( f(2017) \):

\[ f(2016) + f(2017) = \frac{1}{a} \left( (a - 1)^{2016} - 1 \right) + \frac{1}{a} \left( (a - 1)^{2017} + 1 \right) \]

Combining these:

\[ f(2016) + f(2017) = \frac{1}{a} \left( (a - 1)^{2016} + (a - 1)^{2017} \right) \]

### Step 4: Factor out \( (a - 1)^{2016} \)

We can factor out \( (a - 1)^{2016} \):

\[ f(2016) + f(2017) = \frac{(a - 1)^{2016}}{a} \left( 1 + (a - 1) \right) \]

This simplifies to:

\[ f(2016) + f(2017) = \frac{(a - 1)^{2016}}{a} \cdot a = (a - 1)^{2016} \]

### Step 5: Substitute \( a \)

Now substituting \( a = 3^{1/224} + 1 \):

\[ f(2016) + f(2017) = (3^{1/224})^{2016} = 3^{2016/224} = 3^{9} \]

### Step 6: Find \( k \)

Given that \( f(2016) + f(2017) = 3^k \), we have:

\[ 3^k = 3^9 \]

Thus, \( k = 9 \).

### Final Answer

The value of \( k \) is:

\[ \boxed{9} \]