heureka

Thank you CPhill,

here is a nice youtube video

How To prove that 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...))))

Thank you CPhill,

here is a nice youtube video

How To prove that 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...))))

Prove that \(3=\sqrt{1+2\sqrt{1+3 \sqrt{1+4 \sqrt{1+ \cdots}}}}\).

There exist several positive integers x such that
\(\dfrac{1}{x^2 + 2x} \) is a terminating decimal.
What is the second smallest such integer?

Terminating decimal, if \(2,5 | x^2 + 2x\)

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{2} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{2} \\ (x+1)^2 &=& 1 \pmod{2} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{2} \\ x+1 &=& \pm 1 \pmod{2} \\ \hline x+1 &=& +1 \pmod{2} \quad | \quad -1 \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline x+1 &=& -1 \pmod{2} \quad | \quad -1 \\ x &=& -2 \pmod{2} \\ x &=& -2+2 \pmod{2} \\ x &=& 0 \pmod{2} \qquad \text{All even numbers: } x=0,~2,~4,~6,~\dots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \pmod{5} \quad | \quad +1 \\ x^2+2x+1 &=& 1 \pmod{5} \\ (x+1)^2 &=& 1 \pmod{5} \\ \sqrt{ (x+1)^2 } &=& \sqrt{ 1 } \pmod{5} \\ x+1 &=& \pm 1 \pmod{5} \\ \hline x+1 &=& +1 \pmod{5} \quad | \quad -1 \\ x &=& 0 \pmod{5} \qquad x= 0,~5,~10,~15,~\dots \\ \hline x+1 &=& -1 \pmod{5} \quad | \quad -1 \\ x &=& -2 \pmod{5} \\ x &=& -2+5 \pmod{5} \\ x &=& 3 \pmod{5} \qquad x=3+5n,~ n \in \mathbb{Z} \\ \hline \end{array}\)

positive integers \(x= 2,~ 3,~ 4,~ 5,~ 6,~ 8,~ \dots\)

The second smallest such integer is 3

if z is a complex number such that \(z+\dfrac{1}{z}= \sqrt{3}\)
what is the value of \(z^{2010}+\dfrac{1}{z^{2010}}\)?

1. \(|z| = ~?\)

\(\begin{array}{|rcll|} \hline z+\dfrac{1}{z}&=& \sqrt{3} \quad | \quad \text{Let $z=a+bi$} \\\\ a+bi+\dfrac{1}{a+bi}&=& \sqrt{3} \\\\ a+bi+\dfrac{1}{(a+bi)}*\dfrac{(a-bi)}{(a-bi)}&=& \sqrt{3}\quad | \quad (a+bi)*(a-bi)=|z|^2 \\\\ a+bi+\dfrac{a-bi}{|z|^2}&=&\sqrt{3} \\\\ \underbrace{ a+\dfrac{a}{|z|^2}}_{=\sqrt{3}}+ \underbrace{ bi-\dfrac{bi}{|z|^2} }_{=0}&=&\sqrt{3} \\\\ \hline bi-\dfrac{bi}{|z|^2} &=& 0 \\\\ bi &=& \dfrac{bi}{|z|^2} \\\\ |z|^2 &=& \dfrac{bi}{bi} \\\\ \mathbf{ |z|^2 } &=& \mathbf{1} \quad \text{or} \quad \mathbf{|z|=1} \\ \hline \end{array}\)

2. \(\varphi= ~?\)

\(\begin{array}{|rcll|} \hline z&=& |z|*e^{i\varphi} \quad | \quad |z|=1 \\ z&=& e^{i\varphi} \\ \mathbf{z}&=& \mathbf{\cos(\varphi)+i*\sin(\varphi)} \\ \hline \dfrac{1}{z}&=&z^{-1} \\ &=& \left( |z|*e^{i\varphi} \right)^{-1} \\ &=& |z|^{-1}* \left(e^{i\varphi}\right)^{-1} \\ &=& \dfrac{1}{|z|}*e^{i\varphi(-1)} \\ &=& \dfrac{1}{|z|}*e^{i(-\varphi} \quad | \quad |z|=1 \\ &=& e^{i(-\varphi} \\ &=& \cos(-\varphi)+i*\sin(-\varphi) \\ \mathbf{\dfrac{1}{z}} &=& \mathbf{\cos(\varphi)-i*\sin(\varphi)} \\ \hline z+\dfrac{1}{z} &=& \cos(\varphi)+i*\sin(\varphi)+\cos(\varphi)-i*\sin(\varphi) \\\\ z+\dfrac{1}{z} &=& 2\cos(\varphi) \quad | \quad z+\dfrac{1}{z} = \sqrt{3} \\ \sqrt{3} &=& 2\cos(\varphi) \\ \cos(\varphi) &=& \dfrac{\sqrt{3}}{2} \\\\ \mathbf{\varphi} &=& \mathbf{30^\circ} \\ \hline \end{array}\)

3. \(z^{2010}=~?\)

\(\begin{array}{|rcll|} \hline z^{2010} &=& \left( |z|*e^{i\varphi} \right)^{2010} \\ &=& |z|^{2010}* \left(e^{i\varphi}\right)^{2010} \\ &=& 1^{2010}*e^{i*2010\varphi} \\ &=& e^{i*2010\varphi} \\ &=& \cos(2010\varphi)+i*\sin(2010\varphi) \quad | \quad \varphi=30^\circ\\ &=& \cos(2010*30^\circ)+i*\sin(2010*30^\circ)\\ &=& \cos(180^\circ)+i*\sin(180^\circ)\\ \mathbf{z^{2010}}&=& \mathbf{-1} \\ \hline \end{array}\)

4. \(\dfrac{1}{z^{2010}}=~?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{z^{2010}} &=& \left( |z|*e^{i\varphi} \right)^{-2010} \\ &=& |z|^{-2010}* \left(e^{i\varphi}\right)^{-2010} \\ &=& 1^{-2010}* e^{i*(-2010\varphi)} \\ &=& e^{i*(-2010\varphi)} \\ &=& \cos(-2010\varphi)+i*\sin(-2010\varphi) \\ &=& \cos(2010\varphi)-i*\sin(2010\varphi) \quad | \quad \varphi=30^\circ\\ &=& \cos(2010*30^\circ)-i*\sin(2010*30^\circ) \\ &=& \cos(180^\circ)-i*\sin(180^\circ) \\ \mathbf{\dfrac{1}{z^{2010}}}&=& \mathbf{-1} \\ \hline \end{array}\)

\(z^{2010} + \dfrac{1}{z^{2010}} = -1+(-1) \\ z^{2010} + \dfrac{1}{z^{2010}} = -2 \)

The roots of \(x^2 + 5x + 3 = 0\) are p and q.
Find \(\dfrac1p + \dfrac1q\).

\(\text{Set $x=\dfrac{1}{x}$ in $x^2 + 5x + 3$ }\)

\(\begin{array}{|rcll|} \hline x^2 + 5x + 3 &=& 0 \\\\ \Rightarrow \left(\dfrac{1}{x}\right)^2 + 5*\dfrac{1}{x} + 3 &=& 0 \\\\ \dfrac{1}{x^2} + \dfrac{5}{x} + 3 &=& 0 \quad | \quad * x^2 \\\\ 1 + 5x + 3x^2 &=& 0 \\\\ 3x^2+5x+1 &=& 0 \quad | \quad : 3 \\\\ x^2+\dfrac{5}{3}x +\dfrac13 &=& 0 \\ \hline \text{By Vieta:}~\dfrac1p + \dfrac1q &=& -\dfrac{5}{3} \\\\ \text{By Vieta:}~\dfrac1p * \dfrac1q &=& \dfrac{1}{3} \\ \hline \end{array}\)

The teacher has written an equation of the form  on the board,
where  is a quadratic, but Heather can't read the linear term.
She can see that the quadratic term is  4x^2 and that the constant is -24.
She asks her neighbor, Noel, what the linear term is.
Noel decides to tease her and just says, ``One of the roots is 3.''

Heather then says, ``Oh, thanks!''

She then correctly writes down the linear term.
What was the linear term? Include entire term, not simply the coefficient.

\(\begin{array}{|rcll|} \hline 4x^2+ax-24&=& 0 \\ \hline \dfrac{-24}{4} &=& r_1*r_2 \quad | \quad r_1 = 3 \\\\ \dfrac{-24}{4} &=& 3*r_2 \\\\ r_2 &=& \dfrac{-24}{12} \\\\ \mathbf{r_2} &=& \mathbf{-2} \\ \hline \dfrac{a}{4} &=& - (r_1+r_2) \\\\ \dfrac{a}{4} &=& - (3-2) \\\\ \mathbf{ a } &=& \mathbf{ -4 } \\ \hline \end{array}\)

\(4x^2-4x-24\)

Evaluate the sum
\(\dfrac{1}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3} + \dfrac{2}{3^4} +\dfrac{1}{3^5} + \dfrac{2}{3^6} + \dots\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3} + \dfrac{2}{3^4} +\dfrac{1}{3^5} + \dfrac{2}{3^6} + \dots \\\\ &=& \dfrac{1}{3}*\dfrac{3}{3} + \dfrac{2}{3^2} +\dfrac{1}{3^3}*\dfrac{3}{3} + \dfrac{2}{3^4} +\dfrac{1}{3^5}*\dfrac{3}{3} + \dfrac{2}{3^6} + \dots \\\\ &=& \dfrac{5}{3^2} + \dfrac{5}{3^4} + \dfrac{5}{3^6} + \dots \qquad \text{GP:}~ a=\dfrac{5}{3^2},~ r = \dfrac{1}{3^2} \\\\ s&=& \dfrac{5}{3^2} *\left( \dfrac{1}{1-\dfrac{1}{3^2}}\right) \\\\ s&=& \dfrac{5}{9} * \dfrac{9}{8} \\\\ \mathbf{s}&=& \mathbf{\dfrac{5}{8}} \\ \hline \end{array}\)

There are two pairs \((x,y)\) of real numbers that satisfy the equation \(x + y = 4xy = 5\).
Given that the solutions x are in the form \(x = \dfrac{a \pm b \sqrt{c}}{d}\) ,
where a, b, c, and d are positive integers and the expression is completely simplified,
what is the value of \(a + b + c + d\)?

My attempt:

\(\text{Let $a=x+y=5$ } \\ \text{Let $b=xy=\dfrac{5}{4}$ }\)

\(\begin{array}{|rcll|} \hline x^2 - Ax + B &=& 0 \\ x^2 - 5x + \dfrac{5}{4} &=& 0 \\\\ x &=& \dfrac{ 5 \pm \sqrt{25-4\dfrac{5}{4}} }{2} \\\\ x &=& \dfrac{5 \pm \sqrt{25-5} }{2} \\\ x &=& \dfrac{5 \pm \sqrt{20} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{5 \pm 2\sqrt{5} }{2}} \\ \hline x+y &=& 5 \\ y &=& 5-x \\ y &=& 5-\dfrac{5 \pm 2\sqrt{5} }{2} \\\ y &=& \dfrac{10-5 \mp 2\sqrt{5} }{2} \\\\ \mathbf{y} &=& \mathbf{\dfrac{5 \mp 2\sqrt{5} }{2}} \\ \hline \end{array}\)

Two pairs (x,y):
\(\begin{array}{|rcll|} \hline (x_1,~y_1) &=& \left(\dfrac{5 + 2\sqrt{5} }{2},~\dfrac{5 - 2\sqrt{5} }{2} \right) \\\\ (x_2,~y_2) &=& \left(\dfrac{5 - 2\sqrt{5} }{2},~\dfrac{5 + 2\sqrt{5} }{2} \right) \\ \hline \end{array}\)

\(a=5,~ b=2,~ c=5,~ d=2 \\ \mathbf{a+b+c+d = 14}\)

If you were to pick two students out of a class of 7 girls and 6 boys,

what is the probability that the students chosen are two boys?

\(\begin{array}{|rcll|} \hline \dfrac{ \binom{6}{2}_{\text{Boys}}\binom{7}{0}_{\text{Girls} } } {\binom{13}{2}_{\text{Boys and Girls} } } &=& \dfrac{5}{26}\\ \text{or} \\ \dfrac{6}{13}*\dfrac{5}{12}&=& \dfrac{5}{26}\\ \hline \end{array}\)

The probability is \(19.23 \%\)