# Find $( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1$ given ...

For positive real numbers $$a$$ and $$b,$$ let $$a \diamond b = \frac{\sqrt{a^2 + 4ab + b^2 - 2a - 2b + 9}}{ab + 6}.$$Find $$( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1.$$

I can't find any quick way to do this. Can anyone help? Thanks in advance!

• There's nothing other than more parentheses in that leading ellipsis, right? Commented Jun 1, 2020 at 21:26
• yes, there are only the opening parentheses of all the *functions* Commented Jun 1, 2020 at 21:33
• Similar to my question PSE: puzzling.stackexchange.com/questions/99236/… Commented Jul 19, 2020 at 12:04

Notice for all positive number $$a$$, we have

$$a \diamond 2 = \frac{\sqrt{a^2 + 8a + 4 - 2a - 4 + 9}}{2a+6} = \frac{\sqrt{a^2+6a+9}}{2a+6} = \frac12$$

The mess at hand equals to $$( (\cdots) \diamond 2) \diamond 1 = \frac12 \diamond 1 = \frac{\sqrt{37}}{13}$$

• does it follow from the fact the $((a \diamond b) \diamond c) = (a \diamond (b \diamond c))$?
– Alex
Commented Jun 1, 2020 at 22:25
• @Alex, nope. The first thing I do is check whether the $\diamond$ operation is associative and it isn't. Commented Jun 1, 2020 at 22:26
• In your calculation, $a \diamond 2$, what is $a$? All terms between 2017 and 3 or 3?
– Alex
Commented Jun 1, 2020 at 22:33
• @Alex $a$ can be any positive number. if you substitute it by all the terms between 2017 and 3, you get the mess between 2017 and 2. Commented Jun 1, 2020 at 22:36
• OK, so in you construct $a=(\ldots)$, i.e. $\diamond$ between all terms between 3 and 2017. Is it guaranteed to be positive?
– Alex
Commented Jun 1, 2020 at 22:47