### Home > A2C > Chapter 9 > Lesson 9.2.2 > Problem 9-92

Consider this geometric sequence: *i*^{0}, *i*^{1}, *i*^{2}, *i*^{3}, *i*^{4}, *i ^{5} *, …,

*i*

^{15}.

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*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 on 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}= 1What is

*i*^{4}^{n}^{ }, where*n*is a positive whole number?4

*n*is a multiple of 4.Based on your answer to part (c), simplify

*i*^{4}^{n}^{+1},*i*^{4}^{n}^{+2}, and*i*^{4}^{n}^{+3}.Refer to (a), (b), and (c).

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}= ?