The value of sum `sum_(n=1)^(13) ( i^(n) + i^(n+1))` where `i= sqrt( -1)` , is equal to

Options

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

Answer

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

\[ \sum_{n=1}^{13} (i^n + i^{n+1}) \]

where \(i = \sqrt{-1}\).

### Step 1: Simplify the Summation
We can rewrite the summation as follows:

\[ \sum_{n=1}^{13} (i^n + i^{n+1}) = \sum_{n=1}^{13} i^n + \sum_{n=1}^{13} i^{n+1} \]

The second summation can be adjusted by changing the index:

\[ \sum_{n=1}^{13} i^{n+1} = \sum_{m=2}^{14} i^m \]

where \(m = n + 1\). Therefore, we can rewrite the original summation as:

\[ \sum_{n=1}^{13} (i^n + i^{n+1}) = \sum_{n=1}^{13} i^n + \sum_{m=2}^{14} i^m = \sum_{n=1}^{13} i^n + \sum_{n=1}^{13} i^{n+1} - i^1 \]

This results in:

\[ \sum_{n=1}^{13} (i^n + i^{n+1}) = \sum_{n=1}^{14} i^n - i^1 \]

### Step 2: Evaluate the Summation
Now we need to evaluate \(\sum_{n=1}^{14} i^n\). The powers of \(i\) cycle every 4 terms:

- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

Thus, the sum of one complete cycle (4 terms) is:

\[ i + (-1) + (-i) + 1 = 0 \]

Since \(14\) is \(3\) complete cycles plus \(2\) additional terms, we have:

\[ \sum_{n=1}^{12} i^n = 3 \cdot 0 = 0 \]

Now we add the contributions from the last two terms:

\[ \sum_{n=13}^{14} i^n = i^{13} + i^{14} = i + 1 \]

Thus:

\[ \sum_{n=1}^{14} i^n = 0 + (i + 1) = i + 1 \]

### Step 3: Final Calculation
Now we substitute back into our expression:

\[ \sum_{n=1}^{13} (i^n + i^{n+1}) = (i + 1) - i = 1 \]

### Conclusion
The value of the summation is:

\[ \boxed{1} \]