How can we approach a seemingly senseless binomial expansion problem?

[anemone]

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?

 

[I like Serena]

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]

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

 

[anemone]

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. 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.:)

 

[CellCoree]

Hi,

Thank you for sharing your observations and attempt at solving this problem. Binomial expansions can often be tricky and require a lot of careful observation and analysis to arrive at the correct solution.

First, I would like to clarify that the purpose of asking seemingly "senseless" problems under a challenging problems section is to challenge your critical thinking skills and to push you to think outside the box. These types of problems may not have any practical application, but they help to strengthen your problem-solving abilities.

Now, let's address your approach to the problem. It seems like you are on the right track by looking for patterns in the expansion. However, instead of looking at lower values of the exponent, try looking at the higher values. For example, what happens when n=101? Can you see any patterns in the expansion? Also, keep in mind that the value of p can vary depending on the value of b. So, instead of trying to find a specific value for p, try to find a range of possible values.

I hope this helps. Keep exploring and trying different approaches, and you will eventually arrive at the correct answer. Remember, as a scientist, it is important to be persistent and open-minded when faced with challenging problems. Good luck!