The value of ` 1 + sum _(k = 0)^( 12) { cos "" (( 2 k + 1) pi)/( 13) + i sin "" (( 2 k + 1) pi)/( 13)}` is

Options

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

Answer

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

\[ 1 + \sum_{k=0}^{12} \left( \cos\left(\frac{(2k + 1)\pi}{13}\right) + i \sin\left(\frac{(2k + 1)\pi}{13}\right) \right) \]

This expression can be rewritten using Euler's formula:

\[ \cos x + i \sin x = e^{ix} \]

Thus, we can express the summation as:

\[ 1 + \sum_{k=0}^{12} e^{i \frac{(2k + 1)\pi}{13}} \]

### Step 1: Rewrite the Summation
The summation can be rewritten as:

\[ 1 + \sum_{k=0}^{12} e^{i \frac{(2k + 1)\pi}{13}} = 1 + \sum_{k=0}^{12} e^{i \frac{2k\pi}{13}} e^{i \frac{\pi}{13}} \]

This can be factored out as:

\[ 1 + e^{i \frac{\pi}{13}} \sum_{k=0}^{12} \left(e^{i \frac{2\pi}{13}}\right)^k \]

### Step 2: Identify the Geometric Series
The inner summation is a geometric series with first term \(a = 1\) and common ratio \(r = e^{i \frac{2\pi}{13}}\). The number of terms \(n = 13\) (from \(k=0\) to \(k=12\)).

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

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

Substituting the values we have:

\[ \sum_{k=0}^{12} \left(e^{i \frac{2\pi}{13}}\right)^k = \frac{1(1 - (e^{i \frac{2\pi}{13}})^{13})}{1 - e^{i \frac{2\pi}{13}}} \]

### Step 3: Simplify the Exponent
Since \((e^{i \frac{2\pi}{13}})^{13} = e^{i 2\pi} = 1\), we can simplify the sum:

\[ \sum_{k=0}^{12} \left(e^{i \frac{2\pi}{13}}\right)^k = \frac{1 - 1}{1 - e^{i \frac{2\pi}{13}}} = 0 \]

### Step 4: Combine the Results
Now substituting back into our expression:

\[ 1 + e^{i \frac{\pi}{13}} \cdot 0 = 1 + 0 = 1 \]

### Final Answer
Thus, the value of the original expression is:

\[ \boxed{1} \]