Hi Steven, please note the parentheses. [imath](2*3)[/imath] is evaluated first, then its result will * with [imath]4[/imath]. Think of it as composite functions. Let me know if you need further clarification.
Hi Steven, please note the parentheses. [imath](2*3)[/imath] is evaluated first, then its result will * with [imath]4[/imath]. Think of it as composite functions. Let me know if you need further clarification.
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.
I suspect BBB has made a typo in post#13's spoiler
[math]\dfrac{1021615}{1021616}[/math]
I've noticed that...
[math]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)}[/math]
And I'm fairly sure this is easily extended to work for 4 or 6 terms. The latter is shown below...
[math]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)}[/math]
Does this work for any even number of terms? I haven't managed to prove this yet.
[math]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}[/math]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...
[math]x*y=\dfrac{(x+1)(y+1) - (x-1)(y-1)}{(x+1)(y+1) + (x-1)(y-1)}[/math]
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(I strongly suspect that BBB made a typo actually 