heureka

Thank you, CalculatorUser !

Find the number of real solutions to \(x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb\).

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb} \\ \\ x &=& \underbrace{ 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb }_{\text{Infinite Geometric Series}} \\ && a=1,\ r= \dfrac{-x}{1}=\dfrac{x^2}{-x}=\dfrac{-x^3}{x^2}=\dotsb=-x \\ \\ && \boxed{sum = \dfrac{a}{1-r} \text{ as long as } -1 < r < 1 } \\ \\ && sum = \dfrac{1}{1-(-x)} \\ && sum = \dfrac{1}{1+x } \\\\ x &=& \dfrac{1}{1+x } \\ x(1+x) &=& 1 \\ x+x^2 &=& 1 \\ x^2+x-1 &=& 0 \\\\ x &=& \dfrac{-1\pm\sqrt{1^2-4\cdot (-1)}} {2} \\ &=& \dfrac{-1\pm\sqrt{5}} {2} \\\\ \mathbf{x_1} &=& \mathbf{ \dfrac{-1+\sqrt{5}} {2}} \\ r &=& -x_1 \\ &=& -\dfrac{-1+\sqrt{5}} {2} \\ &=& \dfrac{1-\sqrt{5}} {2} \\ r &=& -0.61803398875 \quad | \quad -1 < -0.61803398875 < 1 \checkmark \\\\ \mathbf{x_2} &=& \mathbf{ \dfrac{-1-\sqrt{5}} {2}} \\ r &=& -x_2 \\ &=& -\dfrac{-1-\sqrt{5}} {2} \\ &=& \dfrac{ 1+\sqrt{5}} {2} \\ r &=& 1.61803398875 \quad | \quad -1 < 1.61803398875 < 1 \text{ false, no solution!} \\ \hline \end{array}\)

Real solution is \(x=\mathbf{ \dfrac{-1+\sqrt{5}} {2}}\)

B)
What is \(\sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)}\)?

\(\begin{array}{|rcll|} \hline && \mathbf{ \sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)} } \\\\ &=& \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{ k} - \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=4}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=1}^{\infty} \dfrac{1}{ k+3} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{3} \dfrac{1}{ k} \\\\ &=& \dfrac13 \left( \dfrac{1}{ 1}+\dfrac{1}{ 2}+\dfrac{1}{ 3}\right) \\\\ &=& \dfrac13 \left( \dfrac{6+3+2}{ 6} \right) \\\\ &=& \dfrac13 \left( \dfrac{11}{ 6} \right) \\\\ &=& \mathbf{ \dfrac{11}{ 18} } \\ \hline \end{array}\)

A)
There exist real numbers \(A\) and \(B\) so that \(\dfrac{1}{k(k + 3)} = \dfrac{A}{k} + \dfrac{B}{k + 3}\).
for all real numbers \(k\) other than \(0\) and \(-3\). Enter the ordered pair \((A,B)\).

\(\begin{array}{|lrcll|} \hline &\mathbf{\dfrac{1}{k(k + 3)}} &=& \mathbf{\dfrac{A}{k} + \dfrac{B}{k + 3}} \quad | \quad \cdot k(k + 3) \\\\ &1 &=& \dfrac{Ak(k + 3)}{k} + \dfrac{Bk(k + 3)}{k + 3} \\\\ &1 &=& A (k + 3) + Bk \\ \hline k=-3 &1 &=& A (-3 + 3) + B(-3) \\ &1 &=& 0 -3B \\ &1 &=& -3B \\ & \mathbf{B} &=& \mathbf{-\dfrac{1}{3}} \\ \hline k=0 &1 &=& A (0 + 3) + B(0) \\ &1 &=& 3A+ 0 \\ &1 &=&3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array} \)

\(\mathbf{(A,B) = \left(\dfrac{1}{3},\ -\dfrac{1}{3} \right) }\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{k(k + 3)} &=& \dfrac{\frac{1}{3}}{k} + \dfrac{-\frac{1}{3}}{k + 3} \\ &=& \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \\ \hline \end{array} \)

Hallo tommarvoloriddle,

thank you,

i did make the diagram with Microsoft Word !

Help.

\(\begin{array}{|lrcll|} \hline (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline (1)-(2): & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- (a^2-2ab+b^2) \\ & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- a^2+2ab-b^2 \\ & (a+b)^2-(a-b)^2 &=& 4ab \\ & (a+b)^2 &=& 4ab+(a-b)^2 \\ & \mathbf{ a+b } &=& \mathbf{\sqrt{4ab+(a-b)^2}} \quad | \quad ab=45,\quad a-b = 12 \\ & a+b &=& \sqrt{4\cdot 45+12^2} \\ & a+b &=& \sqrt{324} \\ &\mathbf{ a+b} &=& \mathbf{18} \\ \hline \end{array} \)

The sum of the integers is 18

​ Extremely Difficult German Problem

1.

\(\begin{array}{|rcll|} \hline \mathbf{\sin(\alpha)} &=& \mathbf{\dfrac{1}{x-7}} \qquad (1) \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(180^\circ-(90^\circ-\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(90^\circ+\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\cos(\alpha) }{x} \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\\\ \dfrac{ \sqrt{2}}{2(x-7)} &=& \dfrac{\cos(\alpha) }{x} \\\\ \mathbf{\cos(\alpha)} &=& \mathbf{ \dfrac{ x\sqrt{2}}{2(x-7)} } \qquad (2) \\ \hline \end{array} \)

Phythagoras:

\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)

\(\mathbf{x = 24\ cm}\)

How many positive integers less than \(60\) have an even number of positive divisors?

\(\begin{array}{|r|r|l|} \hline n & \text{the number of divisors of }n \\ \hline 1 & 1 & {\color{red}odd} \\\\ 2&2& {\color{green}even} \\ 3&2& {\color{green}even} \\ 4&3& {\color{red}odd} \\ \\ 5&2& {\color{green}even} \\ 6&4& {\color{green}even} \\ 7&2& {\color{green}even} \\ 8&4& {\color{green}even} \\ 9&3& {\color{red}odd} \\ \\ 10&4& {\color{green}even} \\ 11&2& {\color{green}even} \\ 12&6& {\color{green}even} \\ 13&2& {\color{green}even} \\ 14&4& {\color{green}even} \\ 15&4& {\color{green}even} \\ 16&5& {\color{red}odd} \\\\ 17&2& {\color{green}even} \\ 18&6& {\color{green}even} \\ 19&2& {\color{green}even} \\ 20&6& {\color{green}even} \\ 21&4& {\color{green}even} \\ 22&4& {\color{green}even} \\ 23&2& {\color{green}even} \\ 24&8& {\color{green}even} \\ 25&3& {\color{red}odd} \\ \\ 26&4& {\color{green}even} \\ 27&4& {\color{green}even} \\ 28&6& {\color{green}even} \\ 29&2& {\color{green}even} \\ 30&8& {\color{green}even} \\ 31&2& {\color{green}even} \\ 32&6& {\color{green}even} \\ 33&4& {\color{green}even} \\ 34&4& {\color{green}even} \\ 35&4& {\color{green}even} \\ 36&9& {\color{red}odd} \\ \\ 37&2& {\color{green}even} \\ 38&4& {\color{green}even} \\ 39&4& {\color{green}even} \\ 40&8& {\color{green}even} \\ 41&2& {\color{green}even} \\ 42&8& {\color{green}even} \\ 43&2& {\color{green}even} \\ 44&6& {\color{green}even} \\ 45&6& {\color{green}even} \\ 46&4& {\color{green}even} \\ 47&2& {\color{green}even} \\ 48&10& {\color{green}even} \\ 49&3& {\color{red}odd} \\\\ 50&6& {\color{green}even} \\ 51&4& {\color{green}even} \\ 52&6& {\color{green}even} \\ 53&2& {\color{green}even} \\ 54&8& {\color{green}even} \\ 55&4& {\color{green}even} \\ 56&8& {\color{green}even} \\ 57&4& {\color{green}even} \\ 58&4& {\color{green}even} \\ 59&2& {\color{green}even} \\ \hline && \text{sum = } \color{green}52 \\ \hline \end{array} \)

B:
How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?

I assume:
\(\text{ Let $k=5$ distinct balls } \\ \text{ Let $n=3$ identical boxes } \)

Formula:
\(\begin{array}{|rcll|} \hline \text{Distribution of $k$ distinct Balls into $n$ identical Boxes} =\sum \limits_{i=1}^{n}S(k,i) \\ \text{$S(k,n)$, Stirling number of the second kind } \\ \hline \end{array} \)

A:

Mack the bug starts at $(0,0)$ at noon and each minute moves one unit right or one unit up. He is trying to get to the point $(5,7)$.

However, at $(2,3)$ there is a spider that will eat him if he goes through that point. In how many ways can Mack reach $(5,7)$?