Question

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

Answer 1

`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`

Answer 2

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

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