Imaginary unit/Introductory

Problem

Find the sum of $i^1+i^2+\ldots+i^{2006}$

Solution

Let's begin by computing powers of $i$ .

$i^1=\sqrt{-1}$
$i^2=\sqrt{-1}\cdot\sqrt{-1}=-1$
$i^3=-1\cdot i=-i$
$i^4=-i\cdot i=-i^2=-(-1)=1$
$i^5=1\cdot i=i$

We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences $i^1+i^2+\ldots+i^{4k}$ have a sum of zero (k is a natural number). Since $2006=4\cdot501+2$ , the original series sums to the first two terms of the powers of i, which equals $-1+i$ .

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Imaginary unit/Introductory

Problem

Solution