Compute #i+i^2+i^3+\cdots+i^{258}+i^{259}#. ?

2 Answers
Jul 31, 2018

Answer:

#i+i^2+i^3+\cdots+i^258+i^259=-1#

Explanation:

The trick is to know about the basic idea of sequences and series and also knowing how #i# cycles.

The powers of #i# will result in either: #color(red)(i#, #color(blue)(-1#, #color(green)(-i#, or #color(purple)(1#.

We can regroup #i+i^2+i^3+\cdots+i^258+i^259# into these categories.

We know that #color(red)(i)=color(red)(i^5)=color(red)(i^9)# and so on. The same goes for the other powers of #i#.

So:

#i+i^2+i^3+\cdots+i^258+i^259#

#=color(red)((i+i^5+\cdots+i^257))+color(blue)((i^2+i^6+\cdots+i^258))+color(green)((i^3+i^7+\cdots+i^259))+color(purple)((i^4+i^8+\cdots+i^256))#

We know that within each of these groups, every term is the same, so we are just counting how much of these are repeating.

#=color(red)(65(i))+color(blue)(65(i^2))+color(green)(65(i^3))+color(purple)(64(i^4))#

From here on out, it's pretty simple. You just evaluate the expression:

#=color(red)(65(i))+color(blue)(65(-1))+color(green)(65(-i))+color(purple)(64(1))#
#=\cancel(65i)-65\cancel(-65i)+64#
#=-65+64#
#=-1#

Aug 1, 2018

According to the polynomial identity

$$
\sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1}
$$

we have

$$
i+i^2+i^3+\cdots + i^{259} = \frac{i^{260}-1}{i-1}-1
$$

but #i^260 ⁼( i^4)^{65} = 1# hence

$$
i+i^2+i^3+\cdots + i^{259} = -1
$$