### Home > CCA2 > Chapter 8 > Lesson 8.2.2 > Problem8-90

8-90.

Consider this geometric sequence: $i^0, i^1, i^2, i^3, i^4, i^5 ,...,i^{15}$. .

1. You know that $i^0 = 1$, $i^1 = i$, and $i^2 = -1$. Calculate the result for each term up to $i^{15}$, and describe the pattern.

$i^0 = 1$ Can you see the pattern?
$i^1 = i$
$i^2 = -1$
$i^3 = (i)(i^2) = -i$
$i^4 = (i^2)(i^2) = 1$
$i^5 = (i^2)(i^2)(i) = i$

The pattern repeats $1$$i$, $-1$, and $-i$.

2. Use the pattern you found in part (a) to calculate $i^{16}, i^{25}, i^{39},$ and $i^{100}$.

$16$ is a multiple of $4$.
$25$ is one more than a multiple of $4$.
$39$ is one less than a multiple of $4$.
$100$ is a multiple of $4$.

$i^{16} = 1$
$i^{25} = i$
$i^{39} = -i$
$i^{100} = 1$

3. What is $i^{4n}$ , where $n$ is a positive whole number?

$4n$ is a multiple of $4$.

1. Based on your answer to part (c), simplify $i^{4n+1}$, $i^{4n+2}$, and $i^{4n+3}$.

Refer to (a), (b), and (c).

2. Calculate $i^{396}$, $i^{397}$, $i^{398}$, and $i^{399}$.

$i^{396} = i^{4 · 99} = 1$
$i^{397} = i^{4 · 99+1}=?$
$i^{398} = i^{4 · 99 + 2} = ?$
$i^{399} = ?$