Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 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.

share|cite|improve this question
@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
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 2:53

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

share|cite|improve this answer

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.

share|cite|improve this answer

--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, $\alpha^3 = 1$

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

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

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

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

$S = 1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.