Algebra Problem of The Day-9

[BigBeachBanana]

The operation * is defined as follows:$$x*y=\frac{x+y}{1+xy}$$
What is the value of the following expression?

$$\{[(2*3)*4]...\}*2021=?$$

 

[Steven G]

Are you asking what 2*3*...*2021 equals?

 

[BigBeachBanana]

Are you asking what 2*3*...*2021 equals?

Hi Steven, please note the parentheses. $(2*3)$ is evaluated first, then its result will * with $4$. Think of it as composite functions. Let me know if you need further clarification.

 

[Steven G]

Hi Steven, please note the parentheses. $(2*3)$ is evaluated first, then its result will * with $4$. Think of it as composite functions. Let me know if you need further clarification.

The definition of * has the associative property.

 

[Steven G]

Spoiler

$\displaystyle \dfrac{(2021*2022/2-1)+2021!(1/2+1/3+...+1/2021)}{{1+2021*2022/2 *2020*2021/2-(2*2*3/2+3*3*4/2+...+2020*2020*2021/2)}}$

 

[BigBeachBanana]

Spoiler

$\displaystyle \dfrac{(2021*2022/2-1)+2021!(1/2+1/3+...+1/2021)}{{1+2021*2022/2 *2020*2021/2-(2*2*3/2+3*3*4/2+...+2020*2020*2021/2)}}$

Unfortunately, that's not the correct answer.
Tip: Write out the first few iterations and observe the pattern (if you did, look at the pattern again).

 

[Steven G]

Unfortunately, that's not the correct answer.
Tip: Write out the first few iterations and observe the pattern (if you did, look at the pattern again).

Hmm, I did write out the 1st 5 or 6 iterations and did not observe a pattern. I'll look again!

 

[Steven G]

Spoiler

It seems that the terms are approaching 1

 

[BigBeachBanana]

Hmm, I did write out the 1st 5 or 6 iterations and did not observe a pattern. I'll look again!

Spoiler: Hint

Look at every other iteration.

 

[Steven G]

Spoiler: Hint

Look at every other iteration.

Trust me I have.

 

[Steven G]

Spoiler

Here is what I found:
* is both associative and commutative
2*3*4*5*...*2021 = (2*4)*(6*8)*...*(2018*2020)*(3*5)*(7*9)*...(2019*2021)

Note : (a)*(a+2) = 2(a+1)/(a^2 + 2a + 1) = 2/(a+1). notice that a+1 is the number between

Now (2*4)*(6*8)*...(2018*2020)*(3*5)*(7*9)*...*(2019*2021) = (2/3)*(2/7)*(2/11)*...*(2/2020)
I need some more time to finish up

 

[Steven G]

Spoiler

Some further work.

(2/a)*[2/(a+4)] = 4/(a+2)

So (2*4)*(6*8)*...(2018*2020)*(3*5)*(7*9)*...*(2019*2021) = (2/3)*(2/7)*(2/11)*...*(2/2020)
=(4/5)*(4/13)*(4/21)*...*(4/2014)*(2/2020)

We can continue: (4/a)*(4/(a+8)) = 8/(a+4)...

 

[BigBeachBanana]

Spoiler

Some further work.

(2/a)*[2/(a+4)] = 4/(a+2)

So (2*4)*(6*8)*...(2018*2020)*(3*5)*(7*9)*...*(2019*2021) = (2/3)*(2/7)*(2/11)*...*(2/2020)
=(4/5)*(4/13)*(4/21)*...*(4/2014)*(2/2020)

We can continue: (4/a)*(4/(a+8)) = 8/(a+4)...

It looks like you're making some progress and have put quite a lot of work into this. Here's the answer to keep you going. I'll post the solution in 3 days.

Spoiler: Answer

$$\frac{1\,021\,065}{1\,021\,066}$$

Your response#8 was correct.

 

[Cubist]

I suspect BBB has made a typo in post#13's spoiler

Spoiler: is this the correct answer?

$$\dfrac{1021615}{1021616}$$

Spoiler: observation/ conjecture without answer

I've noticed that...
$$a*b=\dfrac{a+b}{1+ab}=\dfrac{c_1-1}{c_1+1}\\ \text{where }c_1=\dfrac{(a+1)(b+1)}{(a-1)(b-1)}$$
And I'm fairly sure this is easily extended to work for 4 or 6 terms. The latter is shown below...
$$a*b*c*d*e*f=\dfrac{c_2-1}{c_2+1}\\ \text{where }c_2=\dfrac{(a+1)(b+1)(c+1)(d+1)(e+1)(f+1)}{(a-1)(b-1)(c-1)(d-1)(e-1)(f-1)}$$
Does this work for any even number of terms? I haven't managed to prove this yet.

Spoiler: using the above to obtain an answer to the op

$$c=\dfrac{3 \times 4 \times 5 \times \cdots \times 2021\times 2022}{1 \times 2 \times 3 \times \cdots \times 2019\times 2020} = \dfrac{2021\times 2022}{2}=2043231\\ \text{leading to the answer }\dfrac{c-1}{c+1} = \dfrac{1021615}{1021616} \text{ in lowest terms}$$This is slightly different to BBB's answer in post#13 (I strongly suspect that BBB made a typo actually )

EDIT: It took me a very long time to "notice" this! I started out by finding a*b*c*d and removing all the horrible nested fractions. This gave me a clue, when combined with an old thread that I remembered, that initially led me to...
$$x*y=\dfrac{(x+1)(y+1) - (x-1)(y-1)}{(x+1)(y+1) + (x-1)(y-1)}$$

 

[blamocur]

The operation * is defined as follows:$$x*y=\frac{x+y}{1+xy}$$
What is the value of the following expression?

$$\{[(2*3)*4]...\}*2021=?$$

Can I do $1\star 2 \star 3 \star ... \star 2001$ instead ?

 

[blamocur]

1*2*3*...*2021= 1*(2*3*...*2021) = 2*3...*2021 since 1 is the identity number.

I believe '1' works for this operation more like '0' for regular multiplication, i.e. $\forall x: 1\star x = 1$

 

[blamocur]

Spoiler

$$\frac{2021 * 2022 -2}{2021*2022+2}$$

 

[BigBeachBanana]

Spoiler

$$\frac{2021 * 2022 -2}{2021*2022+2}$$

Nice! Well done.

Can I do $1\star 2 \star 3 \star ... \star 2001$ instead ?

Looks interesting. I haven't thought about that. Wondering what we get...

 

[Cubist]

Spoiler

$$\frac{2021 * 2022 -2}{2021*2022+2}$$

Spoiler: and the next step is...

$$= \dfrac{ \left( \dfrac{2021+2022}{1+2021 \times 2022} \right) -2}{ \left( \dfrac{2021+2022}{1+2021 \times 2022}\right) +2}$$just joking

 

[Steven G]

Nice! Well done.


Looks interesting. I haven't thought about that. Wondering what we get...

See post 17