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

2 Answers
Jul 31, 2018

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