Devastating blow

Algebra Level 3

$\large x^{2011} + a_{1}x^{2010} + a_{2} x^{2009} +\ldots + a_{2010} +1$

Given $x_1 , x_2 , x_3 , \ldots , x_{2011}$ are the roots of the polynomial above and that $\begin{aligned} y_1 &=& x_{1} \ x_{2} \cdots x_{10} \\ y_2 &=& x_{2} \ x_{3} \cdots x_{11} \\ y_3 &=& x_{3} \ x_{4} \cdots x_{12} \\ &\cdot& \\ &\cdot& \\ &\cdot& \\ y_{2011} &=& x_{2011}\ x_{1} \ x_{2} \cdots x_{9} \end{aligned}$

Find the value of ${{(y_1 \ y_2\ldots y_{2011})}^{2011}}^{2010}$ .

1 2011 0 None of these -1

1 solution

Marco Giustetto
Jun 9, 2015

First, notice that $a_{2011}$ (the known term) is 1, so the product of all the roots of the polinomial is -1, because of the Vieta's Formula; then, looking at ${{(y_1 \ y_2\ldots y_{2011})}^{2011}}^{2010}$ , we can notice that it is simply the product of all the roots raised to 10, and thus equal to $(-1)^{10} = 1$ .:

indeed, if we expand the writing above using the given definition of the y's (if you need an exact one take a look at the reports), every root appears 10 times (more precisely, every $x_k$ appears in $y_k$ , $y_{k-1}$ , $y_{k-2}$ , $\cdots$ , $y_{k-10}$ ; if the index $i$ of the y is negative, assume $2011+i$ as the index). This implies that

${{(y_1 \ y_2\ldots y_{2011})}^{2011}}^{2010} = \left( \left( \prod x_i \right) ^ {10} \right) ^ { 2011 ^ { 2010} } = {1^{2011}}^{2010} = \boxed 1$

By Vieta's formulas, the product of all the roots of the polynomial is $(-1)^{2011}\frac 11=(-1)$ and not $1$ .

The main idea of your solution is correct that the product of all the $y_i$ 's is the product of all the $x_i$ 's raised to $10$ .

Hence, the required expression is $((-1)^{10})^{2011^{2010}}=1$ .

Please edit your solution accordingly.

Prasun Biswas - 6 years ago

Yes, you're right... Excuse me for the mistake and thank you for the report!

Marco Giustetto - 6 years ago

I can see that you still have a mistake in your work.

${{(y_1 \ y_2\ldots y_{2011})}^{2011}}^{2010} = {{-1}^{2011}}^{2010} = \boxed 1$

I don't get how this is true! Can you see your mistake?

Prasun Biswas - 6 years ago

@Prasun Biswas – Yes, I can... In fact, I was considering $(({-1})^{2010})^{2011}$ and so the mistake I made (that derives from the fact that when you input LaTex you have to input {{-1}^{2011}}^{2010} to make ${{-1}^{2011}}^{2010}$ appear, wich I confused with the writing above); the problem is, when you input 1 as the solution it says that the answer is correct! I'll made a report on this

Marco Giustetto - 6 years ago

@Marco Giustetto – Actually, we have,

$y_1y_2\cdots y_{2011}=(x_1x_2\cdots x_{2011})^{10}=(-1)^{10}$

Hence, you should have,

$\large (y_1 \ y_2\ldots y_{2011})^{2011^{2010}}=((-1)^{10})^{2011^{2010}}=1^{2011^{2010}}=\boxed 1$

I was referring to this. You got the correct answer but your method of evaluating it seems incorrect to me.

Prasun Biswas - 6 years ago

@Prasun Biswas – Ok I've understood... Now I can see my solution has already been edited and seems to be correct, thank you for all the corrections :-)

Marco Giustetto - 6 years ago

Devastating blow

1 solution

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