heureka

Let a and b be real numbers, where |a| < 1and |b| < 1. In an infinite grid, I write the numbers 1, a, a^2, a^3
in the first row. After that, each number is equal to b times the number above it. For example, the numbers in the second row are b, ab, a^2b, a^3b.

Here's the rest of the grid:

(a) Find the sum of all the numbers on the infinite grid.
Infinite sum:

\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right) \)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right)} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \sum \limits_{k=0}^{\infty} a^k \right) \quad | \quad \sum \limits_{k=0}^{\infty} a^k = \dfrac{1}{1-a} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \dfrac{1}{1-a} \right) \\ &=& \left( \dfrac{1}{1-a} \right) \sum \limits_{l=0}^{\infty} b^l \quad | \quad \sum \limits_{l=0}^{\infty} b^l = \dfrac{1}{1-b} \\ &=& \left( \dfrac{1}{1-a} \right) \left( \dfrac{1}{1-b} \right) \\ &=& \mathbf{\dfrac{1}{(1-a)(1-b)}} \\ \hline \end{array}\)

Solve for real numbers x, y, z:
x + yz = 6,
y + xz = 6,
z + xy = 6.

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{x + yz} &=& \mathbf{6} \\ & yz &=& 6-x \\ & \mathbf{z} &=& \mathbf{ \dfrac{6-x}{y} } \quad | \quad \boxed{ y\neq 0 !} \\ \hline (2) & \mathbf{y + xz} &=& \mathbf{6} \\ & xz &=& 6-y \\ & x\left(\dfrac{6-x}{y}\right) &=& 6-y \\ & x(6-x) &=& y(6-y) \\ & 6x-x^2 &=& 6y-y^2 \\ & \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ \hline (3) & \mathbf{z + xy} &=& \mathbf{6} \qquad \text{ or } \qquad \mathbf{z = 6-xy} \\ & xy &=& 6-z \\ & xy &=& 6-\left(\dfrac{6-x}{y}\right) \\ & xy &=& \dfrac{6y-(6-x)}{y} \\ & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ x &=& \dfrac{ 6\pm \sqrt{36-4(6y-y^2)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4y^2-24y+36} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y^2-6y+9)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y-3)^2} } {2} \\ x &=& \dfrac{ 6\pm 2(y-3) } {2} \\ x &=& 3 \pm (y-3) \\ \\ \mathbf{ x = y } \quad &\text{or}& \quad \mathbf{x=6-y} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline x_1 = y: & y^3 &=& 6y-6+y \\ & y^3-7y+6 &=& 0 \\ & y_1 &=& -3 \qquad \Rightarrow x_1 = y_1 = -3 \\ & y_2 &=& 1 \qquad \Rightarrow x_2 = y_2 = 1 \\ & y_3 &=& 2 \qquad \Rightarrow x_3 = y_3 = 2 \\ \hline x_2 = 6-y: & (6-y)y^2 &=& 6y-6+(6-y) \\ & 6y^2-y^3 &=& 5y \\ & y^3- 6y^2 +5y &=& 0 \\ & \underbrace{y}_{y\neq 0} \underbrace{(y^2- 6y +5)}_{=0} &=& 0 \\\\ & y^2- 6y +5 &=& 0 \\ & (y-5)(y-1) &=& 0 \\ & y_4 &=& 5 \qquad \Rightarrow x_4 = 6-y_4 = 1 \\ & y_5 &=& 1 \qquad \Rightarrow x_5 = 6-y_5 = 5 \\ \hline \end{array}\)

\(\begin{array}{|c|r|r|r|} \hline & x & y & z= 6-xy \\ \hline 1. & -3 & -3 & -3 \\ 2. & 1 & 1 & 5 \\ 3. & 2 & 2 & 2 \\ 4. & 1 & 5 & 1 \\ 5. & 5 & 1 & 1 \\ \hline \end{array}\)

Saturn has a radius of about 9.0 earth radii, and a mass 95 times the Earth’s mass.

Estimate the gravitational field on the surface of Saturn compared to that on the Earth.

Newton's law of universal gravitation : \(F = G \dfrac{Mm}{r^2}\)
Newton's laws of motion : \(F=ma\)

\(\begin{array}{|rcll|} \hline F=ma&=& G \dfrac{Mm}{r^2} \\\\ \mathbf{a} &=& \mathbf{\dfrac{GM}{r^2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_{\text{Earth}} &=& \mathbf{\dfrac{GM_{\text{Earth}}}{R_{\text{Earth}}^2}} \\ a_{\text{Saturn}} &=& \mathbf{\dfrac{GM_{\text{Saturn}}}{R_{\text{Saturn}}^2}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{GM_{\text{Saturn}}}{R_{\text{Saturn}}^2}\above 1pt \dfrac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ M_{\text{Saturn}}R_{\text{Earth}}^2}{R_{\text{Saturn}}^2M_{\text{Earth}}} \quad | \quad R_{\text{Saturn}} = 9R_{\text{Earth}},\ M_{\text{Saturn}}= 95M_{\text{Earth}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ 95M_{\text{Earth}}R_{\text{Earth}}^2}{\left(9R_{\text{Earth}}\right)^2M_{\text{Earth}}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ 95R_{\text{Earth}}^2}{81R_{\text{Earth}}^2 } \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& 1.17283950617 \\ a_{\text{Saturn}} &=& 1.17283950617 *a_{\text{Earth}} \quad | \quad a_{\text{Earth}}= 9.8\ \dfrac{m}{s^2} \\ \mathbf{ a_{\text{Saturn}} } &\approx& \mathbf{ 11.5\ \dfrac{m}{s^2} } \\ \hline \end{array}\)

It is possible to place positive integers into the vacant twenty-one squares of this 5*5 grid

so that the numbers in each row and column form an arithmetic sequence.
Find the number that must occupy the vacant square marked by the \(\color{red}{\heartsuit}\).

\(\begin{array}{| *5{@{}c@{}|}} \hline \qquad & \qquad & \qquad & \textcolor{red}{\heartsuit} & \qquad \\ \hline & 74 & & \qquad & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & & & & \\ \hline \end{array}\)

\(\text{The distance of arithmetic sequence row $1 = d_1$ } \\ \text{The distance of arithmetic sequence row $2 = d_2$ } \\ \text{The distance of arithmetic sequence row $3 = d_3$ } \\ \text{The distance of arithmetic sequence row $4 = d_4$ } \\ \text{The distance of arithmetic sequence row $5 = d_5$ } \\ \text{The distance of arithmetic sequence column $1 = d_6$ } \\ \text{The distance of arithmetic sequence column $2 = d_7$ } \\ \text{The distance of arithmetic sequence column $3 = d_8$ } \\ \text{The distance of arithmetic sequence column $4 = d_9$ } \\ \text{The distance of arithmetic sequence column $5 = d_{10}$ } \)

\(\begin{array}{|c| *5{@{}c@{}|}} \hline \color{blue}\text{distance}& \qquad & \qquad & \qquad & \qquad \\ \color{blue}\text{of the rows }\downarrow \\ \hline & & 4d_6+d_5 & 4d_6+2d_5 & 4d_6+3d_5 & 4d_6+4d_5 \\ \color{blue} d_5 & 4d_6 & = & = &=\mathbf{\textcolor{red}{\heartsuit}}= & = \\ & & d_1+4d_7 & 2d_1+4d_8 & 3d_1+4d_9 & 4d_1+4d_{10}\\ \hline & & 3d_6+d_4 & 3d_6+2d_4 & 3d_6+3d_4 & 3d_6+4d_4 \\ \color{blue} d_4 & 3d_6 & \mathbf{=74=} & = & = & = \\ & & d_1+3d_7 & 2d_1+3d_8 & 3d_1+3d_9 & 4d_1+3d_{10} \\ \hline & & 2d_6+d_3 & 2d_6+2d_3 & 2d_6+3d_3 & 2d_6+4d_3 \\ \color{blue} d_3 & 2d_6 & = & = & = & \mathbf{=186=} \\ & & d_1+2d_7 & 2d_1+2d_8 & 3d_1+2d_9 & 4d_1+2d_{10} \\ \hline & & d_6+d_2 & d_6+2d_2 & d_6+3d_2 & d_6+4d_2 \\ \color{blue} d_2 & d_6 & = & \mathbf{=103=} & = & =\\ & & d_1+d_7 & 2d_1+d_8 & 3d_1+d_9 & 4d_1+d_{10} \\ \hline & & & & & \\ \color{blue} d_1 & 0 & d_1 & 2d_1 & 3d_1 & 4d_1 \\ & & & & & \\ \hline \color{green}\text{distance} & & & & & \\ \color{green}\text{of the columns}\rightarrow & \color{green}d_6 & \color{green}d_7 & \color{green}d_8 & \color{green}d_9 & \color{green}d_{10} \\ \hline \end{array}\)

The equations:

\(\begin{array}{|lrcll|} \hline (1) & 3d_6+d_4 &=& 74 \quad \text{or} \quad \mathbf{d_4 = 74 - 3d_6} \\ (2) & d_1+3d_7 &=& 74 \quad \text{or} \quad \mathbf{3d_7 = 74 - d_1} \\ (3) & d_6+2d_2 &=& 103 \quad \text{or} \quad \mathbf{2d_2 = 103 - d_6} \\ (4) & 2d_1+d_8 &=& 103 \quad \text{or} \quad \mathbf{d_8 = 103 - 2d_1} \\ (5) & 2d_6+4d_3 &=& 186 \quad \text{or} \quad \mathbf{4d_3 = 186 - 2d_6} \\ (6) & 4d_1+2d_{10} &=& 186 \quad \text{or} \quad \mathbf{2d_{10} = 186 - 4d_1} \\\\ (7) & \dfrac{(4d_1+d_{10})+(3d_6+4d_4)}{2} &=& 186 \\ & (4d_1+d_{10})+(3d_6+4d_4) &=& 186*2 \\ & 4d_1+\dfrac{186 - 4d_1}{2}+3d_6+4(74 - 3d_6) &=& 372 \\ & \ldots \\ & \mathbf{9d_6-2d_1} &=& \mathbf{17} \qquad (I.) \\\\ (8) & \dfrac{(d_6+4d_2)+(4d_1+3d_{10})}{2} &=& 186 \\ & (d_6+4d_2)+(4d_1+3d_{10}) &=& 186*2 \\ & d_6+4(\dfrac{103 - d_6}{2})+4d_1+3(\dfrac{186 - 4d_1}{2}) &=& 372 \\ & \ldots \\ & \mathbf{d_6+2d_1} &=& \mathbf{113} \qquad (II.) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (I.) + (II.): & 9d_6+d_6 &=& 17+113 \\ & 10d_6 &=& 130 \\ & \mathbf{d_6} &=& \mathbf{13} \\ \hline & 2d_1 &=& 9d_6 - 17 \\ & 2d_1 &=& 9*13 - 17 \\ & 2d_1 &=& 100 \\ & \mathbf{d_1} &=& \mathbf{50} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline d_7 &=& \dfrac{74-d_1}{3} \\\\ d_7 &=& \dfrac{74-50}{3} \\\\ \mathbf{d_7} &=& \mathbf{8} \\ \hline 4d_6+d_5 &=& d_1+4d_7 \\ 4*13+d_5 &=& 50+4*8 \\ 52+d_5 &=& 82 \\ \mathbf{d_5} &=& \mathbf{30} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\textcolor{red}{\heartsuit}} &=& 4d_6+3d_5 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& 4*13+3*30 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& 52+90 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& \mathbf{142} \\ \hline \end{array}\)

If

\(x^2 + y^2 + z^2 = a^2\) and \(x + y + z = x^3 + y^3 + z^3 = a\), then write \(xyz\) expicitly in terms of \(a\).

1.

\(\begin{array}{|rcll|} \hline (x+y+z)^2 &=& x^2 + y^2 + z^2+2(xy+xz+yz) \quad | \quad x + y + z = a,\ x^2 + y^2 + z^2 = a^2 \\ a^2 &=& a^2+2(xy+xz+yz) \\ 2(xy+xz+yz) &=& 0 \\ \mathbf{xy+xz+yz} &=& \mathbf{0} \\ \hline \end{array} \)

2.

Formula: \(x^3+y^3+z^3 = 3xyz+(x+y+z)\left(x^2+y^2+z^2-(xy+xz+yz)\right)\)

\(\begin{array}{|rcll|} \hline \underbrace{x^3+y^3+z^3}_{=a} &=& 3xyz+\underbrace{(x+y+z)}_{=a}\left(\underbrace{x^2+y^2+z^2}_{=a^2}-\underbrace{(xy+xz+yz)}_{=0}\right) \\ a &=& 3xyz+a ( a^2-0 ) \\ a &=& 3xyz+a^3 \\ 3xyz &=& a-a^3 \\ \mathbf{xyz} &=& \mathbf{ \dfrac{1}{3}(a-a^3) } \\ \hline \end{array} \)

In triange ABC, D is on B so that AD:DB = 1:2, and G is on CD so that CG:GD = 3:2. 

If BG intersects AC at F, find BG:GF.

\(\text{Let $AB=\vec{b}$} \\ \text{Let $AC=\vec{c}$} \\ \text{Let $BF=\vec{f}$} \\ \text{Let $CD=\vec{d}$} \)  

\(\begin{array}{|rcll|} \hline (1-\lambda)\vec{b}+x\vec{f}+(1-\mu)\vec{d} &=& \vec{0} \quad | \quad \vec{d} = \lambda\vec{b} - \vec{c},\ \vec{f} = u\vec{c} - \vec{b} \\ (1-\lambda)\vec{b}+x(u\vec{c} - \vec{b})+(1-\mu)(\lambda\vec{b} - \vec{c}) &=& \vec{0} \\ (1-\lambda)\vec{b}+xu\vec{c} - x\vec{b}+(1-\mu)\lambda\vec{b} - (1-\mu)\vec{c} &=& \vec{0} \\ \vec{b}\left[ (1-\lambda) -x +(1-\mu)\lambda \right] &=& \vec{c}\left[ (1-\mu)-xu \right] \\ \vec{b}\left[\underbrace{ (1-\lambda) -x +(1-\mu)\lambda} _{=0}\right] &=& \vec{c}\left[ \underbrace{ (1-\mu)-xu }_{=0}\right] \\\\ (1-\lambda) -x +(1-\mu)\lambda &=& 0 \\ 1-\lambda -x +\lambda -\mu\lambda &=& 0 \\ 1 -x -\mu\lambda &=& 0 \\ \mathbf{x} &=& \mathbf{1-\mu\lambda} \quad | \quad \mu = \dfrac{3}{5},\ \lambda=\dfrac{1}{3} \\ x &=& 1-\dfrac{3}{5}*\dfrac{1}{3} \\ x &=& 1-\dfrac{1}{5} \\ \mathbf{x} &=& \mathbf{\dfrac{4}{5}} \\\\ 1-x &=& 1- \dfrac{4}{5} \\ \mathbf{1-x} &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{BG}{GF}} \\\\ &=& \dfrac{x}{1-x} \\\\ &=& \dfrac{4}{5}\above 1pt \dfrac{1}{5} \\\\ &=& \dfrac{4}{5}*\dfrac{5}{1} \\\\ &=& \mathbf{\dfrac{4}{1}} \\ \hline \end{array}\)

Find the positive integers a and b such that \(\sqrt{10 + \sqrt{84}} = \sqrt{a} + \sqrt{b}\).

\(\begin{array}{|rcll|} \hline \left(\sqrt{a} + \sqrt{b} \right)^2 &=& (a+b) + 2\sqrt{ab} \\ &=& (a+b) + \sqrt{4ab} \quad | \quad \text{compare with } 10 + \sqrt{84} \\\\ && \boxed{ a+b = 10,\ 4ab = 84 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{4ab} &=& \mathbf{84} \quad &| \quad : 4 \\ ab &=& 21 \\ \mathbf{ b} &=& \mathbf{\dfrac{21}{a}} \\ \hline \mathbf{a+b} &=& \mathbf{10} \\ a+\dfrac{21}{a} &=& 10 \quad &| \quad *a \\ a^2+21 &=& 10a \\ a^2-10a+21 &=& 0 \\\\ a &=& \dfrac{10\pm \sqrt{100-4*21}} {2} \\ &=& \dfrac{10\pm \sqrt{16}} {2} \\ &=& \dfrac{10\pm 4} {2} \\ \mathbf{a} &=& \mathbf{5\pm2} \\ a_1 &=& 7 \qquad b_1 = 10-a_1 = 3 \\ a_2 &=& 3 \qquad b_2 = 10-a_2 = 7 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt{10 + \sqrt{84}} &=& \sqrt{7} + \sqrt{3} \quad &| \quad a=7,\ b=3 \\ \sqrt{10 + \sqrt{84}} &=& \sqrt{3} + \sqrt{7} \quad &| \quad a=3,\ b=7 \\ \hline \end{array}\)

There exist digits \(A\) and \(B\) such that \(30AB5 = 225*n\). 

Find all possible values of \(n\).

\(\mathbf{n_{\text{min}}}\)

\(\begin{array}{|rcll|} \hline \mathbf{n_{\text{min}}} &=& \dfrac{30005}{225} \\ &=& 133.3\bar{5} \\ &=& \mathbf{134} \\ \hline \end{array}\)

\(\mathbf{n_{\text{max}}}\)

\(\begin{array}{|rcll|} \hline \mathbf{n_{\text{max}}} &=& \dfrac{30995}{225} \\ &=& 137.7\bar{5} \\ &=& \mathbf{137} \\ \hline \end{array}\)

\(\begin{array}{|r|r|} \hline n & n\times225 \\ \hline 134 & 30150 \\ \color{red}135 & {\color{red}30}27{\color{red}5} \\ 136 & 30600 \\ \color{red}137 & {\color{red}30}82{\color{red}5} \\ \hline \end{array}\)

The possible values of \(n\) are \(\mathbf{135} \) and \(\mathbf{137}\)

One number is chosen at random from the set {1, 2, 3, ..., 6},
and another number is chosen at random from the set {7, 8, 9, ..., 12}.  
What is the probabilty that the product of the two numbers is even?

\(\begin{array}{|c|c|c|c|c|c|c|} \hline \text{product} & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline 1 & & even & & even& & even \\ \hline 2 & even& even& even& even& even& even \\ \hline 3 & & even & & even& & even \\ \hline 4 & even& even& even& even& even& even \\ \hline 5 & & even & & even& & even \\ \hline 6 & even& even& even& even& even& even \\ \hline \end{array}\)

The probabilty that the product of the two numbers is even is \(\dfrac{27}{36} = \mathbf{\dfrac{3}{4}}\)

If

\(x + \dfrac{1}{x} = \dfrac{1 + \sqrt{5}}{2}\), then find \(x^{2000} + \dfrac{1}{x^{2000} }\).

\(\text{Let Golden ratio $\varphi= \dfrac{1 + \sqrt{5}}{2}$ } \\ \text{Let $\varphi^2= 1+\varphi $ } \)

\(\begin{array}{|rcll|} \hline \mathbf{x + \dfrac{1}{x}} &=& \mathbf{ \varphi } \\ \hline x^{2} + \dfrac{1}{x^{2}} &=& \left(x + \dfrac{1}{x} \right)^2 - 2\\ &=& \varphi^2 - 2 \quad &|\quad \varphi^2= 1+\varphi \\ &=& 1+\varphi - 2 \\ \mathbf{x^{2} + \dfrac{1}{x^{2}}} &=& \mathbf{\varphi - 1} \\ \hline x^{3} + \dfrac{1}{x^{3}} &=& \left(x^{2} + \dfrac{1}{x^{2}}\right)\times \left(x + \dfrac{1}{x}\right) - \left(x + \dfrac{1}{x }\right)\\ &=& (\varphi-1 )\times \varphi - \varphi \\ &=& \varphi^2-\varphi - \varphi \quad &|\quad \varphi^2= 1+\varphi \\ &=& 1+\varphi-2\varphi \\ \mathbf{x^{3} + \dfrac{1}{x^{3}}} &=& \mathbf{1-\varphi} \\ \hline x^{5} + \dfrac{1}{x^{5}} &=& \left(x^{2} + \dfrac{1}{x^{2}}\right)\times \left(x^3 + \dfrac{1}{x^3}\right) - \left(x + \dfrac{1}{x }\right)\\ &=& (\varphi-1 )\times (1-\varphi ) - \varphi \\ &=& \varphi^2-\varphi - \varphi \\ &=& \varphi- \varphi^2 -1 + \varphi- \varphi \\ &=& \varphi- \varphi^2 -1 \quad &|\quad \varphi^2= 1+\varphi \\ &=& \varphi- (1+\varphi ) -1 \\ &=& \varphi- 1 -\varphi -1 \\ \mathbf{x^{5} + \dfrac{1}{x^{5}}} &=& \mathbf{-2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^{10} + \dfrac{1}{x^{10}} &=& \left(x^{5} + \dfrac{1}{x^{5}} \right)^2 - 2\\ &=& \left(-2 \right)^2 - 2\\ &=& 4 - 2\\ \mathbf{x^{10} + \dfrac{1}{x^{10}}} &=& \mathbf{2} \\ \hline x^{20} + \dfrac{1}{x^{20}} &=& \left(x^{10} + \dfrac{1}{x^{10}} \right)^2 - 2\\ &=& 2^2 - 2\\ &=& 4 - 2\\ \mathbf{x^{20} + \dfrac{1}{x^{20}}} &=& \mathbf{2} \\ \hline x^{30} + \dfrac{1}{x^{30}} &=& \left(x^{20} + \dfrac{1}{x^{20}}\right)\times \left(x^{10} + \dfrac{1}{x^{10}}\right) - \left(x^{10} + \dfrac{1}{x^{10} }\right)\\ &=& 2\times 2 - 2 \\ &=& 4 - 2\\ \mathbf{x^{30} + \dfrac{1}{x^{30}}} &=& \mathbf{2} \\ \hline x^{40} + \dfrac{1}{x^{40}} &=& \left(x^{30} + \dfrac{1}{x^{30}}\right)\times \left(x^{10} + \dfrac{1}{x^{10}}\right) - \left(x^{20} + \dfrac{1}{x^{20} }\right)\\ &=& 2\times 2 - 2 \\ &=& 4 - 2\\ \mathbf{x^{40} + \dfrac{1}{x^{40}}} &=& \mathbf{2} \\ \hline x^{50} + \dfrac{1}{x^{50}} &=& \left(x^{40} + \dfrac{1}{x^{40}}\right)\times \left(x^{10} + \dfrac{1}{x^{10}}\right) - \left(x^{30} + \dfrac{1}{x^{30} }\right)\\ &=& 2\times 2 - 2 \\ &=& 4 - 2\\ \mathbf{x^{50} + \dfrac{1}{x^{50}}} &=& \mathbf{2} \\ \hline \ldots \\ \hline \mathbf{x^{2000} + \dfrac{1}{x^{2000}}} &=& \mathbf{2} \\ \hline \end{array}\)