## Monday, April 27, 2015

### Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$

Let $x=a$ be a solution of the equation $x^{2012}-7x+6=0$. Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$.

We're told $x=a$ is a solution of the equation $x^{2012}-7x+6=0$, therefore we have $a^{2012}-7a+6=0$.

It can be rewritten as

$a^{2012}-1-7a+6+1=0$

$a^{2012}-1-7a+7=0$

$(a^{2012}-1)-7(a-1)=0$

Here, if we are not familiar with the factorization of the common expression $x^n-1$, then we will face the hard time to manage this particular problem.

In case you haven't already familiar with it, you must understand it and remember it. If it is so hard to memorize, you have got to go over and over it until you knew it by heart. This identity will only help you and save you from the miserable time struggling with some International Mathematical Olympiad contest problems. It will help you to solve any intriguing and necessarily hard and difficult problem elegantly and neatly!

[MATH]\color{yellow}\bbox[5px,purple]{x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x^2+x+1)}[/MATH]

Let do it with some number of $n$ to see why it makes sense:

$x^1-1=x-1$

$x^2-1=(x-1)(x^1+1)$

$x^3-1=(x-1)(x^2+x+1)$

$x^4-1=(x-1)(x^3+x^2+x+1)$ ...

Now, back to our problem where we stopped at $(a^{2012}-1)-7(a-1)=0$:

Apparently, we can factorize $a^{2012}-1=(a-1)(a^{2011}+a^{2010}+\cdots+a+1)$ so we have:

$a^{2012}-1-7a+6+1=0$

$a^{2012}-1-7a+7=0$

$(a^{2012}-1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1)-7(a-1)=0$

$(a-1)(a^{2011}+a^{2010}+\cdots+a+1-7)=0$

It's obvious now that it is either

[MATH]\color{yellow}\bbox[5px,green]{a-1=0}[/MATH] or [MATH]\color{black}\bbox[5px,pink]{a^{2011}+a^{2010}+\cdots+a+1-7=0}[/MATH]

We can conclude by now that

1.

When we have [MATH]\color{yellow}\bbox[5px,green]{a-1=0\implies a=1}[/MATH]:

$1+a+a^2+\cdots+a^{2011}=1+1(2011)=2012$

2.

When we have [MATH]\color{black}\bbox[5px,pink]{a^{2011}+a^{2010}+\cdots+a+1-7=0}[/MATH]:

This implies $1+a+a^2+\cdots+a^{2011}=7$ and we are done.