quadratic equation

Question

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.

Answer 1

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

Answer 2

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.