Evaluate `sum_(n=1)^(13)(i^n+i^(n+1)),` where `n in Ndot`

Explanation

To evaluate the sum \( \sum_{n=1}^{13} (i^n + i^{n+1}) \), we can break it down step by step.

### Step 1: Rewrite the Sum
We can rewrite the sum as follows:
\[ \sum_{n=1}^{13} (i^n + i^{n+1}) = \sum_{n=1}^{13} i^n + \sum_{n=1}^{13} i^{n+1} \]

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