quadratic equation 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. 
 A: To compute quickly using your method, go backwards by $3$'s from $2010$ instead. 
A: 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.
A: --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, $\alpha^3 = 1$
$S = \dfrac {(\alpha^{2011} – 1)} {\alpha - 1}$
$S = \dfrac {(\alpha^{670*3 + 1} – 1)} {\alpha - 1}$
$S = \dfrac {(\alpha^{(3)(670 )}*\alpha) – 1)} {\alpha - 1}$
$S = \dfrac{1*\alpha - 1}{\alpha – 1}$
$S = 1$
