MHB How can we approach a seemingly senseless binomial expansion problem?

AI Thread Summary
The discussion revolves around a binomial expansion problem involving the expression a=(√5+2)^{101}=b+p, where b is an integer and 0<p<1. The initial poster attempts to identify a pattern through lower exponent values but struggles to deduce the answer, questioning the relevance of such problems in the challenging section. Responses clarify that rounding errors led to incorrect calculations, and for higher powers, the terms involving √5 cancel out, resulting in the conclusion that ap=1. The conversation also highlights a connection to the golden ratio, suggesting the problem's design may be inspired by this mathematical concept. Ultimately, the correct evaluation of ap confirms its value as 1.
anemone
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Hi MHB,

I've come across this problem and I think I've observed a pattern when I tried to solve it by using the method of comparison with some lower values of the exponents, but then I just couldn't deduce the answer to the problem because the pattern suggests that I can't. Here is the problem along with my attempt and my question is, am I approaching the problem incorrectly and also, I am wondering what's the point of asking this type of seemingly "senseless" problem under a challenging problems section? (Yes, I found this problem in the challenging problems from a site whose name I don't even recall.)

Problem:

If $$a=(\sqrt{5}+2)^{101}=b+p$$, where $b$ is an integer, $0<p<1$, evaluate $ap$.

Attempt:

Let $$a=(\sqrt{5}+2)^n=b+p$$

[TABLE="class: grid, width: 500"]
[TR]
[TD]$n$[/TD]
[TD]$$a=(\sqrt{5}+2)^n=b+p$$[/TD]
[TD]$ap$[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]$(\sqrt{5}+2)^1$[/TD]
[TD]1.414213562[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]$(\sqrt{5}+2)^3$[/TD]
[TD]0.9999999999999999999999999999999999999999999999999762[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]$(\sqrt{5}+2)^5$[/TD]
[TD]0.9999999999999999999999999999999999999999999999373888[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]$(\sqrt{5}+2)^7$[/TD]
[TD]0.9999999999999999999999999999999999999999999929280362[/TD]
[/TR]
[TR]
[TD]9[/TD]
[TD]$(\sqrt{5}+2)^9$[/TD]
[TD]0.9999999999999999999999999999999999999999956716794758[/TD]
[/TR]
[TR]
[TD]11[/TD]
[TD]$(\sqrt{5}+2)^{11}$[/TD]
[TD]0.9999999999999999999999999999999999886821525305828144[/TD]
[/TR]
[/TABLE]

I noticed that the value of $ap$ deceases at a very small rate and it just is unsafe to say at this point that the value $ap$ that we're looking for in the expansion $$a=(\sqrt{5}+2)^{101}=b+p$$ approaches 1. What do you think?
 
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Hey anemone[/color]! :)

It appears you've made a mistake for n=1:
$$(\sqrt 5 + 2)\{(\sqrt 5 + 2)\} = (\sqrt 5 + 2)(\sqrt 5 - 2) = 1$$

As for your other results, I'd say they are simply 1 instead of 0.9999...
The difference is caused by rounding errors in your calculator.

It appears that for higher powers the $\sqrt 5$ is canceled.
For n=3 we get:
$$(\sqrt 5 + 2)^3\{(\sqrt 5 + 2)^3\} = (17\sqrt 5 + 38)(17\sqrt 5 - 38) = 17^2\cdot 5 - 38^2 = 1$$

What strikes me is the resemblance to the golden ratio number.
$$\varphi = \frac {1+\sqrt 5} {2}$$
$$2+\sqrt 5 = 2\varphi + 1$$
 
If we take

a=(√5+2)^101

and b = (√5-2)^101

and expand both we see that the terms with odd power of (√5) shall be same in both and they shall be positive

so a-b =(√5+2)^101 - (√5-2)^101 is integer

now as (√5-2) < 1 so fractional part of (√5+2)^101 is (√5-2)^101 = p

so ap = (√5+2)^101 * (√5-2)^101 = (5-4) ^ 101 = 1
 
I like Serena said:
Hey anemone[/color]! :)

It appears you've made a mistake for n=1:
$$(\sqrt 5 + 2)\{(\sqrt 5 + 2)\} = (\sqrt 5 + 2)(\sqrt 5 - 2) = 1$$

As for your other results, I'd say they are simply 1 instead of 0.9999...
The difference is caused by rounding errors in your calculator.

It appears that for higher powers the $\sqrt 5$ is canceled.
For n=3 we get:
$$(\sqrt 5 + 2)^3\{(\sqrt 5 + 2)^3\} = (17\sqrt 5 + 38)(17\sqrt 5 - 38) = 17^2\cdot 5 - 38^2 = 1$$

What strikes me is the resemblance to the golden ratio number.
$$\varphi = \frac {1+\sqrt 5} {2}$$
$$2+\sqrt 5 = 2\varphi + 1$$

Thanks for your reply, I like Serena!:)

Oops...you're so right.:o All those values are calculated wrongly as there are rounding errors in the calculations and now I re-do the case for which $n=3$, yes, I get $ap=1$ for that particular case.

Thank you again for spotting my error and hey, now that you mentioned about the golden ratio number, I can tell maybe this is where they got the idea to set this problem up.:)
 
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