Animated Solution for Mathematics - Complex Numbers: The value of the sum ${\sum }_{n=1}^{13}(i^n+i^{n+1})$, where $i=\sqrt{-1}$, equals

The problem asks us to evaluate the sum:

At first glance, it might look like a tedious task of calculating thirteen individual powers of $i$, but we must look for the hidden structure within the expression.

The first step is to simplify the expression inside the summation. We have $i^n+i^{n+1}$, which can be factored as:

Notice that the term $(1+i)$ is completely independent of the summation index $n$. Because $(1+i)$ is a constant relative to the summation, we can pull it outside the sum entirely:

We have effectively separated the constant geometry from the variable rotation.

Now, let us focus on the sum ${\sum }_{n=1}^{13}{i}^n$. The powers of $i$ are not random; they are rotations on the complex plane where $i^1=i$, $i^2=-1$, $i^3=-i$, and $i^4=1$.

These four values repeat in a perfect cycle of length four. The sum of any four consecutive powers of $i$ is:

This is the 'Zero-Sum' property. It is a beautiful cancellation dance where every four steps, the net displacement returns to the origin.

We have thirteen terms in our sum. Since every group of four consecutive terms sums to zero, we can group our thirteen terms into three sets of four, leaving us with just one term at the end.

Specifically, $13=4\times 3+1$. The first twelve terms sum to zero, leaving us with only $i^{13}$. Since:

Our entire summation simplifies to just $i$. Finally, we multiply this result by the constant factor we pulled out earlier:

Since $i^2=-1$, the final result is $i-1$. We have navigated the complexity and arrived at a clean, elegant solution.