heureka

Given that \(\dbinom{15}{8}=6435\), \(\dbinom{16}{9}=11440\) , and \(\dbinom{16}{10}=8008\), find \(\dbinom{15}{10}\).

Formula:  \(\dbinom{n}{k} + \dbinom{n}{k+1} = \dbinom{n+1}{k+1}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\dbinom{n}{k} + \dbinom{n}{k+1}} &=& \mathbf{\dbinom{n+1}{k+1}} \\\\ (1) & \dbinom{15}{8} + \dbinom{15}{9} &=& \dbinom{16}{9} \\ (2) & \dbinom{15}{9} + \dbinom{15}{10} &=& \dbinom{16}{10} \\ \hline (1)-(2): & \dbinom{15}{8} + \dbinom{15}{9} -\left( \dbinom{15}{9} + \dbinom{15}{10}\right) &=& \dbinom{16}{9}-\dbinom{16}{10} \\ & \dbinom{15}{8} - \dbinom{15}{10} &=& \dbinom{16}{9}-\dbinom{16}{10} \\ & 6435 - \dbinom{15}{10} &=& 11440-11440 \\ & 6435 - \dbinom{15}{10} &=& 3432 \\ & \dbinom{15}{10} &=& 6435 - 3432 \\ & \mathbf{\dbinom{15}{10}} &=& \mathbf{3003} \\ \hline \end{array}\)

PSL4#52

The total number of tiles that the diagonals pass through is:

\(\begin{array}{|rcll|} \hline 5\cdot 11 + 8 = 63 \\ \hline \end{array}\)

Thank you, CPhill !

This number \(15^5\) has a long sequence of positive consecutive numbers.
The arithmetic mean of 162nd, 163rd and 164th terms equals 10x, or 10 times, the first term.
What are the first term, last term and
the number of terms of this sequence?

\(\text{Let the first term $=a_1$ } \\ \text{Let the number of terms $=n$ } \\ \text{Let the last term $=a_n$ } \\ \text{Let $a_{163} = a_{162} + 1 $ } \\ \text{Let $a_{164} = a_{162} + 2 $ } \\ \text{Let the sum of the sequence $ 15^5=a_1+a_2+\ldots +a_{n-1}+a_n $ }\)

\(\mathbf{a_1=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{a_{162}+a_{163}+a_{164}} {3} &=& 10a_1 \\ \dfrac{a_{162}+(a_{162} + 1)+(a_{162} + 2)} {3} &=& 10a_1 \\ \dfrac{3a_{162}+3} {3} &=& 10a_1 \\ a_{162}+1 &=& 10a_1 \\ && \mathbf{a_n = a_1+(n-1)d} \quad | \quad d = 1 \\ && \boxed{a_n = a_1+ n-1} \\ && a_{162} = a_1+162-1 \\ && \mathbf{a_{162} = a_1 +161} \\ a_{162}+1 &=& 10a_1 \\ (a_1 +161)+1 &=& 10a_1 \\ a_1 +162 &=& 10a_1 \\ 9a_1 &=& 162 \\ \mathbf{a_1} &=& \mathbf{18} \\ \hline \end{array} \)

\(\mathbf{n=\ ?}\)

\(\begin{array}{|rcll|} \hline 15^5 &=& a_1+a_2+\ldots a_{n-1}+a_n \\ 15^5 &=& a_1+(a_1+1)+(a_1+2)+\ldots +(~a_1+(n-2)~)+(~a_1+(n-1)~) \\ 15^5 &=& na_1+\dfrac{\Big(1+(n-1)\Big)}{2}(n-1) \\ 15^5 &=& na_1+\dfrac{ n(n-1) }{2} \quad | \quad \cdot 2 \\ 2\cdot15^5 &=& 2na_1+ n(n-1) \\ \mathbf{2na_1+ n(n-1)} &=& \mathbf{2\cdot15^5} \quad | \quad a_1 = 18 \\ 2\cdot 18n+ n^2-n &=& 2\cdot15^5 \\ n^2 + 35 n -2\cdot 15^5 &=& 0 \\\\ n &=& \dfrac{-35\pm \sqrt{35^2-4\cdot(-2\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{35^2+8\cdot 15^5)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{1225+6075000)}} {2} \\ n &=& \dfrac{-35\pm \sqrt{6076225 }} {2} \\ n &=& \dfrac{-35\pm 2465} {2} \quad | \quad n>0!\\\\ n &=& \dfrac{-35+ 2465} {2} \\ \mathbf{ n} &=& \mathbf{1215} \\ \hline \end{array}\)

\(\mathbf{a_n=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1+n-1} \\ a_n &=& 18+1215-1 \\ \mathbf{ a_n} &=& \mathbf{1232} \\ \hline \end{array}\)

\(\mathbf{check:}\)

\(\begin{array}{|rcll|} \hline 15^5 &=& 18 + 19 +20 + \ldots +1231+1232 \\ 15^5 &=& \dfrac{(18+1232)}{2}\cdot 1215 \\ 15^5 &=& \dfrac{1250}{2}\cdot 1215 \\ 15^5 &=& 625\cdot 1215 \\ 759375&=& 759375\ \checkmark \\ \hline a_{162} &=& 18 + 162-1 \\ a_{162} &=& 179 \\\\ \dfrac{179+180+181}{3} &=& 10 \cdot 18 \\ \dfrac{540}{3} &=& 180 \\ 180 &=& 180\ \checkmark \\ \hline \end{array}\)

Let \(x,y,z \) be nonnegative real numbers such that \(x + y + z = 3\)

Find the maximum value of \((xy + z)(xz + y)\)

\(\begin{array}{|rcll|} \hline \huge{GM} &\huge{\leq}& \huge{AM} \\\\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{(xy + z)+(xz + y)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ xy + z + xz + y }{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{x(y+z)+(y+z)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (y+z)(x+1)}{2} \quad | \quad \boxed{y + z = 3-x} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \\ &&\begin{array}{|rcll|} \hline \large{GM} &\large{\leq}& \large{AM} \\\\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ (3-x)+(x+1)}{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 3-x + x+1 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 4 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & 2 \quad | \quad \text{square both sides}\\ (3-x)(x+1) & \leq & 4 \\ \mathbf{ \dfrac{(3-x)(x+1)}{2}} & \leq & \mathbf{ 2 } \\ \hline \end{array} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \leq 2 \quad | \quad \text{ The maximum value results from the equals sign} \\ \sqrt{(xy + z)(xz + y)} & = & \dfrac{ (3-x)(x+1)}{2} = 2 \quad | \quad \text{square both sides}\\ \mathbf{(xy + z)(xz + y) } &=& \mathbf{4} \\ \hline \end{array}\)

The maximum value of \(\mathbf{(xy + z)(xz + y)=4}\)

Compute  \(\Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg\rfloor\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg \rfloor }\\ &=& \Bigg\lfloor \dfrac{2007!}{2006! + 2005!} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~(1+2006)} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~2007} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006! }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2005!~2006 }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor 2006 + \underbrace{\dfrac{2004!}{2006! + 2005!}}_{<1} \Bigg \rfloor \\ &=& \mathbf{2006} \\ \hline \end{array}\)

Thank you, Melody !

Find all solutions to the inequality \(\large{\dfrac{(2x-7)(x-3)}{x} \ge 0}\) in interval notation

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{(2x-7)(x-3)}{x}} &\ge& \mathbf{0} \quad | \quad \boxed{x\neq 0 !} \\\\ \dfrac{(2x-7)(x-3)}{x} &\ge& 0 \quad | \quad \times x^2 \\ \dfrac{(2x-7)(x-3)x^2}{x} &\ge& 0\times x^2 \\ (2x-7)(x-3)x &\ge& 0 \\ 2\left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \quad | \quad : 2 \\ \left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \\ \mathbf{ \left(x-\dfrac{7}{2}\right)(x-3)(x-0) } &\ge& \mathbf{0} \\ \hline \end{array}\)

We create a sign table:

\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Interval or position} : & (-\infty,0) & 0 & (0,3) & 3 & \left(3,\dfrac{7}{2}\right) &\dfrac{7}{2} & \left(\dfrac{7}{2},\infty\right) \\ \hline \text{sign of } (x-0): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x-3): & - & - & - & 0 & + & + & + \\ \hline \text{sign of } \left(x-\dfrac{7}{2}\right): & - & - & - & - & - & 0 & + \\ \hline \text{sign of }\left(x-\dfrac{7}{2}\right)(x-3)(x-0): & - & \color{red}0 & \color{red}+ & \color{red}0 & - & \color{red}0 & \color{red}+ \\ \hline \end{array}\)

We can read the result:

in the interval \([0,3] \)and \(\Big[\dfrac{7}{2},\infty\Big)\) the left side of the inequality is positive or zero, and thus the inequality is true there.

But \(\boxed{x\neq 0 !}\) and the solution is: \(\Big(0,3\Big]\) and \(\Big[\dfrac{7}{2},\infty\Big)\)

Potenz- und Winkelfunktionen

Mit Wurzelfunktionen:

\(\begin{array}{|rcll|} \hline d^2 &=& a^2+x^2 \quad | \quad (\text{nehme beide Seiten hoch } \dfrac{3}{2}) \\ \Big(d^2\Big)^{\dfrac{3}{2}} &=& \Big(a^2+x^2\Big)^{\dfrac{3}{2}} \\ \Big(d \Big)^{2\dfrac{3}{2}} &=& \Big(a^2+x^2\Big)^{\dfrac{3}{2}} \\ \mathbf{ d^3 } &=& \mathbf{\Big(a^2+x^2\Big)^{\dfrac{3}{2}} } \\ \hline \end{array} \)

Ersetzen von d durch die Wurzelfunktion:

\(\begin{array}{|rcll|} \hline \mathbf{I} &=& \mathbf{\dfrac{x}{d^3}} \\ \hline I &=& \dfrac{x}{\Big(a^2+x^2\Big)^{\dfrac{3}{2}}} \\\\ \mathbf{I} &=& \mathbf{x\Big(a^2+x^2\Big)^{-\dfrac{3}{2}} } \\ \hline \end{array}\)

\(\mathbf{ I_{\text{max}}=\ ? }\)
Die Ableitung nach x wird 0 gesetzt:

\(\begin{array}{|rcll|} \hline \dfrac{d\ I}{d\ x} &=& 1*\Big(a^2+x^2\Big)^{-\dfrac{3}{2}}+x\left(-\dfrac{3}{2}\right)\Big(a^2+x^2\Big)^{-\dfrac{3}{2}-1}*2x \\ \dfrac{d\ I}{d\ x} &=& \Big(a^2+x^2\Big)^{-\dfrac{3}{2}}-3x^2\Big(a^2+x^2\Big)^{-\dfrac{3}{2}-1} \\ \dfrac{d\ I}{d\ x} &=& \Big(a^2+x^2\Big)^{-\dfrac{3}{2}} -3x^2\Big(a^2+x^2\Big)^{-\dfrac{3}{2}}\Big(a^2+x^2\Big)^{-1} \\\\ && \boxed{\dfrac{d\ I}{d\ \alpha} = 0} \\\\ 0 &=& \Big(a^2+x^2\Big)^{-\dfrac{3}{2}} -3x^2\Big(a^2+x^2\Big)^{-\dfrac{3}{2}}\Big(a^2+x^2\Big)^{-1} \\ 3x^2\Big(a^2+x^2\Big)^{-\dfrac{3}{2}}\Big(a^2+x^2\Big)^{-1} &=& \Big(a^2+x^2\Big)^{-\dfrac{3}{2}} \quad | \quad : \Big(a^2+x^2\Big)^{-\dfrac{3}{2}} \\ 3x^2 \Big(a^2+x^2\Big)^{-1} &=& 1 \quad | \quad * \Big(a^2+x^2\Big)^{1} \\ 3x^2 &=& \Big(a^2+x^2\Big)^{1} \\ 3x^2 &=& a^2+x^2 \quad | \quad - x^2 \\ 2x^2 &=& a^2 \quad | \quad (\text{auf beiden Seiten die Wurzel ziehen}) \\ \sqrt{2}x &=& a \quad | \quad * \sqrt{2} \\ 2x &=& a\sqrt{2} \quad | \quad : 2 \\ x &=& a\dfrac{\sqrt{2}} {2} \quad | \quad a = 8 \\ x &=& 8\dfrac{\sqrt{2}} {2} \\ \mathbf{x} &=& \mathbf{4 \sqrt{2}} \\ \hline \end{array}\)

I hat ein Maximum an der Stelle  \(\mathbf{4\sqrt{2}}\)

Potenz- und Winkelfunktionen

Wenn es sich um eine Extremwertaufgabe handelt, dann ist die Lösung  folgende:

Mit Winkelfunktionen:

\(\begin{array}{|lrcll|} \hline (1)& \tan(\alpha) &=& \dfrac{x}{a} \\ \text{ oder } & x&=&a\tan(\alpha) \\ \hline (2)& \cos(\alpha) &=& \dfrac{a}{d} \\ \text{ oder } & d&=& \dfrac{a}{\cos(\alpha)} \\ \hline \end{array}\)

Ersetzen von x und d durch die Winkelfunktionen:

\(\begin{array}{|rcll|} \hline \mathbf{I} &=& \mathbf{\dfrac{x}{d^3}} \\ \hline I &=& \dfrac{a\tan(\alpha)}{\left( \dfrac{a}{\cos(\alpha)} \right)^3} \\\\ I &=& \dfrac{a\tan(\alpha)}{\left( \dfrac{a^3}{\cos^3(\alpha)}\right) } \\\\ I &=& \dfrac{a\tan(\alpha)\cos^3(\alpha)}{a^3} \\\\ I &=& \dfrac{ \tan(\alpha)\cos^3(\alpha)}{a^2} \\ \\ I &=& \dfrac{1}{a^2}\dfrac{ \sin(\alpha)\cos^3(\alpha)}{\cos(\alpha)} \\ \\ \mathbf{I} &=& \mathbf{\dfrac{1}{a^2} \sin(\alpha)\cos^2(\alpha) } \\ \hline \end{array} \)

\(\mathbf{ I_{\text{max}}=\ ? }\)

Die Ableitung nach \(\alpha\) wird 0 gesetzt:

\(\begin{array}{|rcll|} \hline \dfrac{d\ I}{d\ \alpha} &=& \dfrac{1}{a^2} \left[ \sin(\alpha)* 2 *\cos(\alpha)(-\sin(\alpha)) +\cos(\alpha)\cos^2(\alpha) \right] \\ \dfrac{d\ I}{d\ \alpha} &=& \dfrac{1}{a^2} \left[ \cos^3(\alpha)- 2\sin^2(\alpha)\cos(\alpha) \right] \\\\ && \boxed{\dfrac{d\ I}{d\ \alpha} = 0} \\\\ \dfrac{1}{a^2} \left[ \cos^3(\alpha)- 2\sin^2(\alpha)\cos(\alpha) \right] &=& 0 \quad | \quad *a^2\\ \cos^3(\alpha)- 2\sin^2(\alpha)\cos(\alpha)&=& 0 \\ \mathbf{\cos(\alpha) \left[\cos^2(\alpha)- 2\sin^2(\alpha) \right] }&=& \mathbf{0} \\ \underbrace{\cos(\alpha)}_{\neq 0,\ (\alpha \neq 90^\circ)} \left[\underbrace{\cos^2(\alpha)- 2\sin^2(\alpha)}_{=0} \right]&=& 0 \\\\ \cos^2(\alpha)- 2\sin^2(\alpha) &=& 0 \\ 2\sin^2(\alpha) &=& \cos^2(\alpha)\quad | \quad : \cos^2(\alpha) \\ \\ 2\tan^2(\alpha) &=& 1 \\ \tan^2(\alpha) &=& \dfrac{1}{2} \\ \tan (\alpha) &=& \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\ \mathbf{\tan (\alpha)} &=& \mathbf{\dfrac{\sqrt{2}}{2}} \qquad (\alpha =35.2643896828\ldots ^\circ ) \\ \hline \end{array}\)

I hat ein Maximum an der Stelle \(x =a\tan(\alpha)\)

\(\begin{array}{|rcll|} \hline x &=& a\tan(\alpha) \\ x &=& 8\dfrac{\sqrt{2}}{2} \\ \mathbf{x} &=& \mathbf{4\sqrt{2}} \\ \hline \end{array}\)

I hat ein Maximum an der Stelle  \(\mathbf{x=4\sqrt{2}}\)