Melody

sqrt (3332)

factor(3332) = (2^2*7^2)*17

\(\sqrt{3332}=2*7*\sqrt{17}=14\sqrt{17}\)

Yeah right :/

Now Chris,

You would have seen me working on it before you even started your answer :/

You are one rude guest.  !!

If I am the first to see you demand an answer like this again I will delete your question immediately.

I am only leaving it here now out of respect for the answerer and so that you can see this response!

9^3k * 27^3k-1 * 3^k  all divided by 81^4k  like a fraction

\(\;\;\;\frac{9^{3k} * 27^{3k-1} * 3^k }{ 81^{4k} }\\ =\frac{3^{2*3k} * 3^{3(3k-1)} * 3^k }{ 3^{4*4k} }\\ =\frac{3^{6k} * 3^{9k-3} * 3^k }{ 3^{16k} }\\ =3^{6k+9k-3+k-16k}\\ =3^{-3}\\ =\frac{1}{3^3}\\ =\frac{1}{27}\)

I didn't know that was the Fibonacci sequence!  Thanks Heureka :)

mmm  I think Majid may have said something like that.

It has never happened to me before.   Weird.....

It is probably just the ghost of web2 turning in his grave....

Mmm

\(\dfrac{1}{\sqrt5}\cdot\left(\left(\dfrac{1+\sqrt5}{2}\right)^5-\left(\dfrac{1-\sqrt5}{2}\right)^5\right)\)

\((1+\sqrt5)^5-(1-\sqrt5)^5\\ =1+5(\sqrt5)+10(\sqrt5)^2+10(\sqrt5)^3+5(\sqrt5)^4+(\sqrt5)^5\\ -(1+5(-\sqrt5)+10(-\sqrt5)^2+10(-\sqrt5)^3+5(-\sqrt5)^4+(-\sqrt5)^5)\\~\\ =+5(\sqrt5)+10(\sqrt5)^3+(\sqrt5)^5\\ -(+5(-\sqrt5)+10(-\sqrt5)^3+(-\sqrt5)^5)\\~\\ =10\sqrt5+20(\sqrt5)^3+2(\sqrt5)^5\\ =10\sqrt5+100(\sqrt5)+50(\sqrt5)\\ =160\sqrt5 \)

SO

\(\frac{1}{\sqrt5}\cdot (\frac{160\sqrt5}{32})\\ =\frac{160}{32}\\ =5\)

(you forgot your signature Max LOL)

~ The smartest dinosaur's mum in the world ~   Red is so proud of me :D

Hi Max,

I did reload my page.  I reloaded it a number of times.

It is there now but it took about 7 minutes before it displayed for me.

That has never happened to me before.  VERY STRANGE!

Hi Arshan,  :)

I have only seen this for the first time tonight.

Based on what i watched in this clip