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

Consider this geometric sequence:

*i*^{0},*i*^{1},*i*^{2},*i*^{3},*i*^{4},*i*, …,^{5}*i*^{15}. Homework Help ✎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.Use the pattern you found in part (a) to calculate

*i*^{16},*i*^{25},*i*^{39}, and*i*^{100}.What is

*i*^{4}^{n}^{ }, where*n*is a positive whole number?Based on your answer to part (c), simplify

*i*^{4}^{n}^{+1},*i*^{4}^{n}^{+2}, and*i*^{4}^{n}^{+3}.Calculate

*i*^{396},*i*^{397},*i*^{398}, and*i*^{399}.

*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*.

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

4*n* is a multiple of 4.

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

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