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If $\alpha$ is root of equation $x^2+x+1 = 0$ then find the value of $1+\alpha +\alpha^2+\alpha^3+\cdots+\alpha^{2010}$

Here I have put the value of $\alpha$ in the given equation to get $1+\alpha + \alpha^2$ which is similar to the first three terms. So, each three terms give value = 0 . Only the last term will remain which is $\alpha^{2010}$

Can we equate this with the help of Geometric progression somehow....as the given terms form a G.P with first term 1 and common ratio $\alpha$

Sum of the $n$ terms of G.P $= \dfrac{a(1-r^{n})}{1-r}$ where r is common ratio .

Please suggest.

Michael Hardy
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Sachin Sharmaa
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    @Sashin Sharmaa I see this is your tenth question here. You haven't accepted any answers yet. Please consider accepting your favorite answer to each question if any of the answers was helpful. – Git Gud Feb 20 '13 at 17:16
  • @Sashin Sharmaa Do you understand why "So, each three terms give value $0$"? – Git Gud Feb 20 '13 at 17:16
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    Note that $x^3-1 = (x-1)(x^2+x+1)$, so $\alpha^3 = 1$. – Thomas Andrews Feb 20 '13 at 17:20
  • Hi All, thanks for that... however I got the answer using what you are saying, but my question was that.. can we solve this by using Geometric progression method somehow. Thanks.. – Sachin Sharmaa Feb 21 '13 at 02:53

3 Answers3

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To compute quickly using your method, go backwards by $3$'s from $2010$ instead.

André Nicolas
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If $1 + \alpha + \alpha^2 = 0$, what does that say about $\alpha^1 + \alpha^2 + \alpha^3$? And $\alpha^2 + \alpha^3 + \alpha^4$ and so on?

Once you've figured this out, you can subtract any sequence of three $\alpha$ terms, not just ones that start with an exponent of a multiple of three.

Joe Z.
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--Method 1

In $1+\alpha +\alpha^2+\alpha^3+\cdots+\alpha^{2010}$, there are $2011$ terms $(= 670*3 + 1$ terms).

As mentioned in your work, the sum of any three consecutive terms is 0. Your grouping should then be

$ (\alpha^{2010} + \alpha^{2009} + \alpha^{2008}) + \cdots + (\alpha^3 +\alpha^2+\alpha) + 1$

From which you should get 1 as the result.

--Method 2 (by summation of a geometric progression)

As pointed out already, $\alpha^3 = 1$

$S = \dfrac {(\alpha^{2011} – 1)} {\alpha - 1}$

$S = \dfrac {(\alpha^{670*3 + 1} – 1)} {\alpha - 1}$

$S = \dfrac {(\alpha^{(3)(670 )}*\alpha) – 1)} {\alpha - 1}$

$S = \dfrac{1*\alpha - 1}{\alpha – 1}$

$S = 1$

Mick
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