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18:14 Uhr, 25.07.2022

Tag!
dear supporters:
(3 + 2√2)⁵ < a , and a is natural number. What is the min value for A?.

what math skill is being measured in this exercise. I only want you to tell me the skill. not to solve the problem for me. I will do my research and show you my work after that.
Vielen Dank

18:26 Uhr, 25.07.2022

You need the knowledge about Pascal's triangle and binomial theorem.

18:28 Uhr, 25.07.2022

Thank you so much.

18:37 Uhr, 25.07.2022

de.serlo.org/mathe/1723/pascalsches-dreieck

You can also use your calculator.

The result is: de.serlo.org/mathe/1723/pascalsches-dreieck

If you round: the outcome is $6726$

18:38 Uhr, 25.07.2022

thanks@supporter for the link.

19:37 Uhr, 25.07.2022

To avoid irrational numbers you can calculate

$2\cdot \sum _{k\phantom{\rule{0.12em}{0ex}}=1,3,5}\left(\left(\genfrac{}{}{0ex}{}{5}{k\phantom{\rule{0.12em}{0ex}}}\right)\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot \left(2\sqrt{2}{\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$

19:44 Uhr, 25.07.2022

Ich bezwéifle, dass RASTA mit Summen Erfahrung hat.

19:56 Uhr, 25.07.2022

Das ist ja auch, meiner Meinung nach, keine Schulaufgabe. Rasta postet ja öfters Aufgaben, welche eher jenseits der Schule behandelt werden. Mein Eindruck ist, dass viele Aufgaben nicht aus einem offiziellen Lehrbuch stammen und die Aufgabenschwierigkeit dann natürlich auch oberhalb der Fähigkeiten von Rasta liegen.
Deswegen hatte ich schon einmal die Empfehlung gegeben sich ein offizielles Lehrbuch anzuschaffen.

Wenn man den binomischen Lehrsatz regulär anwendet, dann benötigt man einen Taschenrechner. Dann kann man aber auch gleich bei dem ursprünglichen Term den Taschenrechner verwenden.

20:58 Uhr, 25.07.2022

thank you to the three of you. Yes,I have little experience with this type of problem and summation or sum notation but I want to get practiced in that. Thank you for helping.
i just finished watching a tutorial about Pascal's triangle. very interesting but I fail to see the connection with the problem at hand. anyways, it was interesting.

21:06 Uhr, 25.07.2022

pivot, i have a textbook, in fat i have three textbooks. I saw this problem posted by someone and got curious and wanted to know the skills needed to solve that. I studied the triangle a bit, don't see how to apply it but and i'ma be watching and studying the binomial theorem. I like to do that once in a while go rogue, lol, but thn i will aslk you only for the skill needed and then i may try to do it on my own as a challenge and post some woek which you will surely rectify. just that.
thank you all

21:09 Uhr, 25.07.2022

Questions:
1. What is the source of that exercise?
2. Do you live in Germany?
3. Do you go to school in Germany?
4. If the answer to 2. is yes, then are you planning to buy a Lehrbuch?

21:14 Uhr, 25.07.2022

it is an exercise I saw in a high school test practice online. don;t remember tnow because it was a pdf. My dad travels to Frankfurt and he sometimes takes me. I live Gulf shores Alabama, USA. I have textbooks i am studying 9th grade math now and I will soon begin 10th grade math.

21:26 Uhr, 25.07.2022

(3 + 2√2)5 = 6725.9998..., so the minimum value of A is 6726.
the answer to the exercise was this one but it did not have any steps, just the right answer
I tried to do it with my calculator but somehow was not even getting 6725.... because of course, i was having doubts adding a whole number with a number times a square root. never done that calculation before.
thanks

21:31 Uhr, 25.07.2022

i dont know what k stands for in the equation pivot showed for me.

21:35 Uhr, 25.07.2022

I started watching a video on the binomial theorem and I will take notes and try to finish pivots' equation. Don't tell me anythin, plase, let me show some work of my own. it will take me sometime but I will post at least my effort.

21:46 Uhr, 25.07.2022

supporter, the result you say is in the triangle and you posted the link, plsase, if it is not too much to ask show me how I can find it. I watched a whole video about the triangle but i need to practice at least with one problem and then I will know.
www.youtube.com/watch?v=XMriWTvPXHI

23:07 Uhr, 25.07.2022

Eine mögliche Variante, den Wert $a\phantom{\rule{0.12em}{0ex}}$ OHNE Nutzung eines Taschenrechners zu bestimmen:

${a\phantom{\rule{0.12em}{0ex}}}_{n\phantom{\rule{0.12em}{0ex}}}={\left(3+2\sqrt{2}\right)}^{n\phantom{\rule{0.12em}{0ex}}}+{\left(3-2\sqrt{2}\right)}^{n\phantom{\rule{0.12em}{0ex}}}$ ist die explizite Darstellung der rekursiven Folge ${a\phantom{\rule{0.12em}{0ex}}}_{n\phantom{\rule{0.12em}{0ex}}}=6{a\phantom{\rule{0.12em}{0ex}}}_{n\phantom{\rule{0.12em}{0ex}}-1}-{a\phantom{\rule{0.12em}{0ex}}}_{n\phantom{\rule{0.12em}{0ex}}-2}$ mit den Startwerten ${a\phantom{\rule{0.12em}{0ex}}}_{0}=2$ sowie ${a\phantom{\rule{0.12em}{0ex}}}_{1}=6$. Gemäß dieser Rekursion sind die weiteren Werte

${a\phantom{\rule{0.12em}{0ex}}}_{2}=6\cdot 6-2=34$
${a\phantom{\rule{0.12em}{0ex}}}_{3}=6\cdot 34-6=198$
${a\phantom{\rule{0.12em}{0ex}}}_{4}=6\cdot 198-34=1154$
${a\phantom{\rule{0.12em}{0ex}}}_{5}=6\cdot 1154-198=6726$,

letzteres entspricht wegen $0<{\left(3-2\sqrt{2}\right)}^{5}<1$ dem gesuchten $a\phantom{\rule{0.12em}{0ex}}$.

23:43 Uhr, 25.07.2022

thank you so much,Hal, for taking the time to explain.

23:43 Uhr, 25.07.2022

thank you so much,Hal, for taking the time to explain.

05:55 Uhr, 26.07.2022

@HAL:
Das dürfte RASTA total überfordern.
Aber es ist immer interessant, was einem Profis dazu einfällt.
Wer ko, der ko, sagt man da in Bayern! :-)
Der Laie verstummt und bewundert. Profi ist halt Profi. Chapeau!

08:23 Uhr, 26.07.2022

Die binomische Entwicklung von ${a\phantom{\rule{0.12em}{0ex}}}_{5}$ führt übrigens auf den Term von Pivot, allgemein

${a\phantom{\rule{0.12em}{0ex}}}_{n\phantom{\rule{0.12em}{0ex}}}=2\sum _{r\phantom{\rule{0.12em}{0ex}}=0}^{⌊\frac{n\phantom{\rule{0.12em}{0ex}}}{2}⌋}\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{c}n\phantom{\rule{0.12em}{0ex}}\\ 2r\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{n\phantom{\rule{0.12em}{0ex}}-2r\phantom{\rule{0.12em}{0ex}}}\cdot {\left(2\sqrt{2}\right)}^{2r\phantom{\rule{0.12em}{0ex}}}=2\sum _{r\phantom{\rule{0.12em}{0ex}}=0}^{⌊\frac{n\phantom{\rule{0.12em}{0ex}}}{2}⌋}\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{c}n\phantom{\rule{0.12em}{0ex}}\\ 2r\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{n\phantom{\rule{0.12em}{0ex}}-2r\phantom{\rule{0.12em}{0ex}}}\cdot {8}^{r\phantom{\rule{0.12em}{0ex}}}$ .

Ob man das nun zur Berechnung heranzieht oder doch die Rekursion, ist letztlich Geschmackssache.

09:13 Uhr, 26.07.2022

Wie kommt man darauf mit der Differenz zu arbeiten bzw. der rekursiven Folge?
Welcher Gedankengang steckt dahinter?
Wann braucht man in der Praxis das Pascalsche Dreieck?
Hast du vlt. ein Exempel aus der Wirtschaft oder Technik.
Ich bin immer anwendungsorientiert, so prägt sich vieles bei mir besser ein.

13:25 Uhr, 26.07.2022

If I use of a calculator I find that (3 + 2√2)5 = 6725.9998..., so the minimum value of A is 6726.
But that is t he easy way. i wanted to learn the other way. like you well said the explanations went over my head but they gave me a reference. Now i continue to study pascal's triangle and its relation with coefficients and little by little i will see the light. that is how the first mathematicians learn, by trial and error. nothing wrong with that.

18:05 Uhr, 26.07.2022

I will explain my approach. Suppose you have the term $\left(3+2\sqrt{2}{\right)}^{5}+\left(3-2\sqrt{2}{\right)}^{5}$, where $3-2\sqrt{2}<1$

Now we use the binomial expansion for both parts.

$\sum _{k\phantom{\rule{0.12em}{0ex}}=0}^{5}\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}+\sum _{k\phantom{\rule{0.12em}{0ex}}=0}^{5}\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(-2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$

For $k\phantom{\rule{0.12em}{0ex}}=0,2,4$ the exponent $5-k\phantom{\rule{0.12em}{0ex}}$ is odd. Thus ${\left(-2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$ is negative and therefore $\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(-2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$ negative. So the sum is 0 for $\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}+\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(-2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$ at $k\phantom{\rule{0.12em}{0ex}}=0,2,4$.

Thus the term can be written as
$2\cdot \sum _{k\phantom{\rule{0.12em}{0ex}}=1,3,5}^{5}\left(\phantom{\rule{0.167em}{0ex}}\begin{array}{}5\\ k\phantom{\rule{0.12em}{0ex}}\end{array}\phantom{\rule{0.167em}{0ex}}\right)\cdot {3}^{k\phantom{\rule{0.12em}{0ex}}}\cdot {\left(2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$. This is the upper bound $a\phantom{\rule{0.12em}{0ex}}$.

Since the exponents of 5-k are even for $k\phantom{\rule{0.12em}{0ex}}=1,3,5$ the numbers of ${\left(2\sqrt{2}\right)}^{5-k\phantom{\rule{0.12em}{0ex}}}$ are all integers. So the sum can be calculated without using a calculator. Nevertheless here is the input/output of wolfram alpha.

www.wolframalpha.com/input?i=2*%28%285+choose+1%29*3%5E1*%282*sqrt%282%29%29%5E%285-1%29%2B%285+choose+3%29*3%5E%283%29*%282*sqrt%282%29%29%5E%285-3%29%2B%285+choose+5%29*3%5E%285%29*%282*sqrt%282%29%29%5E%285-5%29%29

Here is a link which explains the sigma sign.

http//www.mathcentre.ac.uk/resources/workbooks/mathcentre/sigma.pdf

20:00 Uhr, 26.07.2022

Thank you so much, pivot,
agift here for you and for the rest fof the supporters. Perhaps you enjoy it.
www.sciencealert.com/7-times-mathematics-became-art-and-blew-our-minds

19:17 Uhr, 27.07.2022

Thanks for the gift. Very beautiful pictures-created by math.