`("^(m)C_(0)+^(m)C_(1)-^(m)C_(2)-^(m)C_(3))+('^(m)C_(4)+^(m)C_(5)-^(m)C_(6)-^(m)C_(7))+..=0` if and only if for some positive integer `k`, `m=` (a) 4k (b) 4k+1 (c) 4k-1 (d) 4k+2

Options

A. `4k`
B. `4k+1`
C. `4k-1`
D. `4k+2`

Explanation

To solve the given problem, we need to analyze the expression:

\[ \left( \binom{m}{0} + \binom{m}{1} - \binom{m}{2} - \binom{m}{3} \right) + \left( \binom{m}{4} + \binom{m}{5} - \binom{m}{6} - \binom{m}{7} \right) + \ldots = 0 \]

We will use the properties of binomial coefficients and complex numbers to derive the conditions under which this expression equals zero.

### Step 1: Consider the Binomial Expansion
We start with the binomial expansion of \( (x + y)^m \) and \( (x - y)^m \):

\[ (x + y)^m = \sum_{k=0}^{m} \binom{m}{k} x^{m-k} y^k \]

\[ (x - y)^m = \sum_{k=0}^{m} \binom{m}{k} x^{m-k} (-y)^k \]

### Step 2: Add and Subtract the Expansions
Let \( x = \cos \theta \) and \( y = i \sin \theta \). Then we have:

\[ (\cos \theta + i \sin \theta)^m = e^{i m \theta} \]
\[ (\cos \theta - i \sin \theta)^m = e^{-i m \theta} \]

Adding these two equations gives:

\[ 2 \cos(m \theta) = \sum_{k=0}^{m} \binom{m}{k} \cos^{m-k} \theta (i \sin \theta)^k + \sum_{k=0}^{m} \binom{m}{k} \cos^{m-k} \theta (-i \sin \theta)^k \]

### Step 3: Simplify the Expression
The terms with odd powers of \( \sin \theta \) will cancel out when adding, while the even powers will remain. Thus, we can express the sum as:

\[ 2 \cos(m \theta) = \sum_{k \text{ even}} \binom{m}{k} \cos^{m-k} \theta (-1)^{k/2} \sin^k \theta \]

### Step 4: Analyze the Given Series
The series in the problem can be interpreted as a combination of terms from the binomial expansion where we group the coefficients based on their parity.

### Step 5: Set the Condition for Zero
For the entire expression to equal zero, we need the coefficients of the series to satisfy certain conditions. This leads to the conclusion that:

\[ m - 1 = 4k \quad \text{for some integer } k \]

### Step 6: Solve for \( m \)
From the equation \( m - 1 = 4k \), we can express \( m \) as:

\[ m = 4k + 1 \]

### Conclusion
Thus, the value of \( m \) that satisfies the original equation is:

\[ \boxed{4k + 1} \]