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

Explanation

To solve the problem, we need to compute the value of

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

### Step 1: Recognize the Summation as a Complex Exponential
Using Euler's formula, we can rewrite the summation:

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

Thus, we have:

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

### Step 2: Express the Summation in Terms of a Geometric Series
The terms in the summation form a geometric series where the first term \( a = e^{i\frac{\pi}{15}} \) and the common ratio \( r = e^{i\frac{2\pi}{15}} \). The number of terms \( n = 15 \).

The sum of a geometric series can be calculated using the formula:

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

Substituting our values:

\[ S = e^{i\frac{\pi}{15}} \frac{1 - \left(e^{i\frac{2\pi}{15}}\right)^{15}}{1 - e^{i\frac{2\pi}{15}}}. \]

### Step 3: Simplify the Expression
Calculating \( \left(e^{i\frac{2\pi}{15}}\right)^{15} \):

\[ \left(e^{i\frac{2\pi}{15}}\right)^{15} = e^{i2\pi} = 1. \]

Thus, the sum simplifies to:

\[ S = e^{i\frac{\pi}{15}} \frac{1 - 1}{1 - e^{i\frac{2\pi}{15}}} = e^{i\frac{\pi}{15}} \cdot 0 = 0. \]

### Step 4: Final Calculation
Now, substituting back into our original expression:

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

### Conclusion
The final value is:

\[ \boxed{1}. \]