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If $a$ is a root of $x^2 + x + 1$ simplify $$1 + a + a^2 + a^3 + \cdots + a^{2017}.$$

my solution initially starts with the idea that $1 + a + a^{2} = 0$ since $a$ is one of the root then using the idea i grouped $$1 + a + a^2 + a^3 + \cdots + a^{2017}$$ just like this $$(1 + a + a^{2}) + a^3(1 + a + a^2) + a^6(1 + a + a^2) + \cdots + a^{2013} + (1 + a + a^2) + a^{2016} + a^{2017}$$ which is equivalent to $$(0) + a^3(0) + a^6(0) + \cdots + a^{2013}(0) + a^{2016} + a^{2017}$$ and finally simplified into $$a^{2016} + a^{2017}$$. That's my first solution, but eventually I noticed that since $1 + a + a^2 = 0$ then it can be $a + a^2 = -1 $ using this idea I further simplified $$a^{2016} + a^{2017}$$ into $$a^{2015}(a + a^{2})$$ ==> $$a^{2015}(-1)$$ ==> $$-a^{2015}$$ But answers vary if I start the grouping at the end of the expression resulting to $1 + a$ another solution yields to $-a^{2017}$ if I set $1 + a = -a^2$ several solutions rise when the grouping is being made anywhere at the body of the expression... my question is, Is $$a^{2016} + a^{2017} = 1 + a = -a^{2015} = -a^{2017} \text{ etc.?}$$ How does it happen? I already forgot how to manipulate complex solutions maybe that's why I'm boggled with this question...any help?

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    $\begingroup$ $ a^3 = 1 $ .......... $\endgroup$ – Will Jagy Aug 28 '17 at 1:33
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    $\begingroup$ @rosa A typo in your question: $a^{2016} + a^{2017} = 1 + a = -a^{2015} = -a^{2}$? $\endgroup$ – Math Lover Aug 28 '17 at 1:43
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    $\begingroup$ $x^2+x+1$ has the following roots: $x = -(-1)^(1/3) or (-1)^(2/3) = a$ therefore what Will said is true: $a^3 = 1$ $\endgroup$ – user29418 Aug 28 '17 at 1:45
  • $\begingroup$ if I start the grouping at the end of the expression resulting to 1+a That's a perfectly good way to obtain the correct answer. $\endgroup$ – dxiv Aug 28 '17 at 2:43
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I haven't looked through your working, but here's a much easier method:

If $a^2+a+1=0$, then $(a-1)(a^2+a+1)=0$, that is, $a^3-1=0$, ie, $a^3=1$.

On the other hand, $1+a+\dots+a^{2017}=\frac{a^{2018}-1}{a-1}$. Since $a^3=1$, it follows that $a^{2018}=a^2$. So, your sum simplifies to $\frac{a^2}{a-1}$. <-- wrong! Read on....

Edit:

I had a typo (thanks @Math lover for pointing that out). The sum simplifies to $\frac{a^2-1}{a-1}$, which simplifies easily to $a+1$.

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    $\begingroup$ There is a typo. The sum is $\frac{a^2-1}{a-1} = a+1=-a^2$. $\endgroup$ – Math Lover Aug 28 '17 at 1:40
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    $\begingroup$ Thank a lot for clarifications :) $\endgroup$ – rosa Aug 28 '17 at 1:53
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You are so close to figuring this out on your own!

Here's a "meta"-question and some ideas for you.

Meta-question: What do think the purpose of this exercise is? Other than busy work, of course. What do you think will be learned?

ideas: 1) If $a$ is a root to $1 + x + x^2$ then as $1+x+x^2$ is a quadratic there are two possible values for $a$ which we can figure out via the quadratic equation. Because $b^2 - 4ac = (1)^2 - 4(1)(1) = - 3< 0$ neither of these values are real.

It could be illuminating to actually figure out the values of $a$, to graph them on the complex plane, and to figure out what $a^k$ would be. But that's not actually the point of the problem.

2) If $a$ is a root to $1 + x + x^2$, then $a$ is also a root to $(1-x)(1 + x + x^2) = x^3 -1$.

$x^3 -1$ is a third degree polynomial and has $3$ roots. One of them is the real $x = 1$. The other non-real two roots are the two possible values of $a$.

And that is the answer to the meta-question. $a$ is what is called a third root of unity. The polynomial $x^k - 1=0$ will have $k$ complex roots. $x=1$ will be one of them. If $k$ is even $x = -1$ will be another. The rest will be complex and they will be roots of the polyomial $1 + x + x^2 + ..... + x^{k-1}$.

There will be $n-1$ such roots and they will all but so that $x^k = 1$.

3) Another way of thinking about 2) above:

$1 + a + a^2 = 0$

$a(1 + a + a^2) = 0*a = 0$

$a + a^2 + a^3 = 0$

$1 + a + a^2 + a^3 = 1 + 0 = 1$

$(1 + a + a^2) + a^3 = 1$

$a^3 = 1$.

So $1 + a + a^2 + a^3 + ....... + a^{2017} = $

$(1+a + a^2)(1 + a^3 + a^6 + ....... + a^{2013}) + a^{2016} + a^{2017}=$

$0*(1+1+........ + 1) + a^{2016}(1 + a) = $

$1(1+a)= 1 + a$

=== post script ====

Okay. Back to 1)...

$a^2 + a + 1 = 0$

$a = \frac {-1 \pm \sqrt{-3}}2 = -\frac 12 \pm i\frac {\sqrt{3}}2$.

(This should look familiar. If $x = -\frac 12$ and $y=\pm \frac {\sqrt{3}}2$ what kind of angle is being formed by $(x,y)$ and $(0,0)$ and the $x$-axis?)

(Food for thought: What kind of shape is made be the three points $a, -a$ and $1$. i.e. what shape is made by the three third roots of unity?)

(What shape do you think will be made by the $k$ $k$-roots of unity?)

An $1 + a + .... +a^{2017} = 1 + a = \frac 12 \pm i\frac {\sqrt{3}}2$.

One thing very much worth noting:

$a^0 = 1$

$a^1 = a = -\frac 12 \pm i\frac {\sqrt{3}}2$

$a^2 = (-\frac 12 \pm i\frac {\sqrt{3}}2)^2 =$

$\frac 14 - \frac 34 \pm i*2(-\frac 12)(\frac{\sqrt{3}}2) =$

$= -\frac 12 \mp i\frac {\sqrt{3}}2 = -a$.

$a^3 = (-\frac 12 \pm i\frac {\sqrt{3}}2)(-\frac 12 \mp \frac {\sqrt{3}}2) =$

$= \frac 14 - (-\frac 34) = \frac 14 + \frac 34 = 1$.

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  • $\begingroup$ This is a full lecture nice one :) thanks a lot, I notice that you consider all possible consideration in different bodies of math :) nice note :) $\endgroup$ – rosa Aug 29 '17 at 12:36

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