The value of
`1+sum_(k=0)^(14) {cos((2k+1)pi)/(15) - isin((2k+1)pi)/(15)}`, is

Options

A. 0
B. -1
C. 1
D. i

Answer

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

\[ 1 + \sum_{k=0}^{14} \left( \cos\left(\frac{(2k+1)\pi}{15}\right) - i \sin\left(\frac{(2k+1)\pi}{15}\right) \right) \]

### Step 1: Rewrite using Euler's formula
We can use Euler's formula, which states that:

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

Thus, we can rewrite the expression as:

\[ \cos\left(\frac{(2k+1)\pi}{15}\right) - i \sin\left(\frac{(2k+1)\pi}{15}\right) = e^{-i\frac{(2k+1)\pi}{15}} \]

This allows us to rewrite the summation:

\[ 1 + \sum_{k=0}^{14} e^{-i\frac{(2k+1)\pi}{15}} \]

### Step 2: Recognize the summation as a geometric series
The summation can be expressed as:

\[ 1 + \sum_{k=0}^{14} e^{-i\frac{(2k+1)\pi}{15}} = 1 + \sum_{k=0}^{14} \left( e^{-i\frac{\pi}{15}} \right)^{(2k+1)} \]

Let \( \alpha = e^{-i\frac{\pi}{15}} \). The summation becomes:

\[ 1 + \sum_{k=0}^{14} \alpha^{(2k+1)} \]

### Step 3: Simplify the summation
The series can be simplified to:

\[ 1 + \alpha + \alpha^3 + \alpha^5 + \ldots + \alpha^{29} \]

This is a geometric series with the first term \( a = \alpha \) and common ratio \( r = \alpha^2 \), with \( n = 15 \) terms.

### Step 4: Use the formula for the sum of a geometric series
The sum of a geometric series is given by:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

In our case, we have:

\[ S = \alpha \frac{1 - (\alpha^2)^{15}}{1 - \alpha^2} \]

Calculating \( (\alpha^2)^{15} = e^{-i\frac{30\pi}{15}} = e^{-i2\pi} = 1 \):

\[ S = \alpha \frac{1 - 1}{1 - \alpha^2} = 0 \]

### Step 5: Combine the results
Thus, we have:

\[ 1 + S = 1 + 0 = 1 \]

### Final Answer
The value of the original expression is:

\[ \boxed{1} \]