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A binomial expansion problem

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,718
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 \(\displaystyle a=(\sqrt{5}+2)^{101}=b+p\), where $b$ is an integer, $0<p<1$, evaluate $ap$.

Attempt:

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

$n$\(\displaystyle a=(\sqrt{5}+2)^n=b+p\)$ap$
1$(\sqrt{5}+2)^1$1.414213562
3$(\sqrt{5}+2)^3$0.9999999999999999999999999999999999999999999999999762
5$(\sqrt{5}+2)^5$0.9999999999999999999999999999999999999999999999373888
7$(\sqrt{5}+2)^7$0.9999999999999999999999999999999999999999999929280362
9$(\sqrt{5}+2)^9$0.9999999999999999999999999999999999999999956716794758
11$(\sqrt{5}+2)^{11}$0.9999999999999999999999999999999999886821525305828144

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 \(\displaystyle a=(\sqrt{5}+2)^{101}=b+p\) approaches 1. What do you think?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,796
Hey anemone! :)

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.
$$\varphi = \frac {1+\sqrt 5} {2}$$
$$2+\sqrt 5 = 2\varphi + 1$$
 

kaliprasad

Well-known member
Mar 31, 2013
1,311
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
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,718
Hey anemone! :)

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.:eek: 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.:)