Today I continued the math book I was doing, and I did problems 1-9 in Chapter 2 :)
My favorite problem was an AIME problem: 
My Solution:
Note that, by the Binomial Theorem, $$x^{2001} + (\frac{1}{2} - x)^{2001}$$ $$= x^{2001} + \left(-x^{2001} + \binom{2001}{1}x^{2000} \left(\frac{1}{2}\right)^{1} - \binom{2001}{2}x^{1999}\left(\frac{1}{2}\right)^{2} + \dots + \binom{2001}{2001}x^0 \left(\frac{1}{2}\right)^{2001} \right)$$ The $x^{2001}$ terms cancel, and we know that this whole expression is $0$, so $$=\binom{2001}{1}x^{2000} \left(\frac{1}{2}\right)^{1} - \binom{2001}{2}x^{1999}\left(\frac{1}{2}\right)^{2} + \dots + \binom{2001}{2001}x^0 \left(\frac{1}{2}\right)^{2001} = 0$$ Now, we have to find the sum of roots of this expression. For this, we can use Vieta's formula for the sum of roots: $$\sum \text{roots} = -\frac{-\binom{2001}{2} \cdot \frac{1}{4}}{\binom{2001}{1} \cdot \frac{1}{2}}=\boxed{500}.$$
Cool solution by the author:
Note that when you plug in $x=\frac{1}{2} - r$, then you get the same equation as when you plug in $x=r$.
This implies that the solutions to this equation come in pairs: if $r$ is a solution, then $\frac{1}{2} - r$ is also a solution. Since there are pairs of $(r, \frac{1}{2}-r)$, the sum of all of the roots would be $\frac{1}{2}$ summed up several times. It would be summed for each pair, and since we have a polynomial of degree $2000$, there are $$\frac{2000}{2}=1000$$ pairs. The sum of roots is thus $\frac{1}{2} \cdot 1000 = \boxed{500}.$
I think the author's solution was a cool and slick way that was applicable in this problem, but in my opinion, the best way to approach these types of problems with a foolproof approach is the Vieta's formulas.
ok
ReplyDeletefr
ReplyDelete