`underset(r=0)overset(n)(sum)(-1)^(r).^(n)C_(r)[(1)/(2^(r))+(3^(r))/(2^(2r))+(7^(r))/(2^(3r))+(15^(r))/(2^(4r))+ . . .m" terms"]=`

Options

A. `(2^(mn)-1)/(2^(mn)(2^(n)-1))`
B. `(2^(mn)-1)/(2^(n)-1)`
C. `(2^(mn)+1)/(2^(n)+1)`
D. `(2^(mn)+1)/(2^(n)-1)`

Answer

To solve the given summation problem, we will follow a systematic approach using the Binomial Theorem and properties of series. The expression we need to evaluate is:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \ldots \text{ (m terms)} \right) \]

### Step 1: Identify the pattern in the series
The series inside the summation can be rewritten based on the pattern of the coefficients. The coefficients appear to follow the form \(2^r - 1\) for each term. Thus, we can express the series as:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( \frac{2^r - 1}{2^r} + \frac{4^r - 1}{4^r} + \frac{8^r - 1}{8^r} + \ldots \right) \]

### Step 2: Rewrite the series
We can rewrite the series as follows:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( \frac{2^r - 1}{2^r} + \frac{4^r - 1}{4^r} + \frac{8^r - 1}{8^r} + \ldots \right) = \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( 1 - \frac{1}{2^r} + 1 - \frac{1}{4^r} + 1 - \frac{1}{8^r} + \ldots \right) \]

### Step 3: Factor out common terms
We can factor out the common terms from the summation:

\[ = \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( m - \left( \frac{1}{2^r} + \frac{1}{4^r} + \frac{1}{8^r} + \ldots \right) \right) \]

### Step 4: Use the Binomial Theorem
Using the Binomial Theorem, we can evaluate the summation:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} x^r = (1 - x)^n \]

### Step 5: Substitute values
Now, we substitute \(x = \frac{1}{2}\), \(x = \frac{1}{4}\), \(x = \frac{1}{8}\), etc., into the Binomial expansion:

\[ (1 - \frac{1}{2})^n + (1 - \frac{1}{4})^n + (1 - \frac{1}{8})^n + \ldots \]

### Step 6: Evaluate the series
Evaluating these terms gives us:

\[ \sum_{k=1}^{m} (1 - \frac{1}{2^k})^n \]

### Step 7: Final simplification
Now, we can simplify the result to find the final answer. The result will depend on the number of terms \(m\) and the value of \(n\).

### Final Result
The final result of the summation can be expressed in terms of \(n\) and \(m\) as:

\[ = \sum_{k=1}^{m} (1 - \frac{1}{2^k})^n \]