• Feb 20th 2013, 08:16 AM
sachinrajsharma
If $\displaystyle \alpha$ is root of equation $\displaystyle x^2+x+1 = 0$ then find the value of $\displaystyle 1+\alpha +\alpha^2+\alpha^3+.....+\alpha^{2010}$

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

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

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

• Feb 20th 2013, 08:33 AM
BobP
The equation $\displaystyle \alpha^{2}+\alpha+1=0$ has complex roots, so can't you simply write (them) in exponential form and substitute into $\displaystyle \alpha^{2010}$ ?
• Feb 20th 2013, 08:58 AM
MINOANMAN
SachinRajSharma

The solution is in front of you. let the 1 be alone and combine 3 by 3 the remaining terms that give you zero as per your initial condition.
The answer is 1. as simple as such. If however you want to take and conbine the remaining terms as you said you will get a^2010. find the complex number that represents a and use de Moivre's theorem to rise to the power 2010 the complex number a the result is again 1 but is is lengthy.
just give it a try.

MINOAS