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

Jul 31, 2018

#### Answer:

$i + {i}^{2} + {i}^{3} + \setminus \cdots + {i}^{258} + {i}^{259} = - 1$

#### 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 $i + {i}^{2} + {i}^{3} + \setminus \cdots + {i}^{258} + {i}^{259}$ into these categories.

We know that $\textcolor{red}{i} = \textcolor{red}{{i}^{5}} = \textcolor{red}{{i}^{9}}$ and so on. The same goes for the other powers of $i$.

So:

$i + {i}^{2} + {i}^{3} + \setminus \cdots + {i}^{258} + {i}^{259}$

$= \textcolor{red}{\left(i + {i}^{5} + \setminus \cdots + {i}^{257}\right)} + \textcolor{b l u e}{\left({i}^{2} + {i}^{6} + \setminus \cdots + {i}^{258}\right)} + \textcolor{g r e e n}{\left({i}^{3} + {i}^{7} + \setminus \cdots + {i}^{259}\right)} + \textcolor{p u r p \le}{\left({i}^{4} + {i}^{8} + \setminus \cdots + {i}^{256}\right)}$

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

$= \textcolor{red}{65 \left(i\right)} + \textcolor{b l u e}{65 \left({i}^{2}\right)} + \textcolor{g r e e n}{65 \left({i}^{3}\right)} + \textcolor{p u r p \le}{64 \left({i}^{4}\right)}$

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

$= \textcolor{red}{65 \left(i\right)} + \textcolor{b l u e}{65 \left(- 1\right)} + \textcolor{g r e e n}{65 \left(- i\right)} + \textcolor{p u r p \le}{64 \left(1\right)}$
$= \setminus \cancel{65 i} - 65 \setminus \cancel{- 65 i} + 64$
$= - 65 + 64$
$= - 1$

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

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